Cc chuyn

LUYN THI I HC

Bin son: Nguyn Minh Hiu

THPT Phan nh Phng

ng HiThng 08 - 2012

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xF1 F2A1 A2

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Copyright c2012 by Nguyn Minh Hiu, All rights reserved.

Nguyn Minh Hiu

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Chuyn 1. Kho St S Bin Thin V V Th Hm S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51. Tnh n iu Ca Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52. Cc Tr Ca Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63. Gi Tr Ln Nht V Gi Tr Nh Nht Ca Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74. ng Tim Cn Ca Th Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85. Kho St S Bin Thin V V Th Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Chuyn 2. Phng Trnh - Bt Phng Trnh & H Phng Trnh i S . . . . . . . . . . . . . . . . . 111. Phng Trnh & Bt Phng Trnh Khng Cha Cn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112. Phng Trnh & Bt Phng Trnh Cha Cn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123. H Phng Trnh i S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144. Phng Trnh & H Phng Trnh Cha Tham S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Chuyn 3. Phng Php Ta Trong Mt Phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171. Ta Trong Mt Phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172. Phng Trnh ng Thng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183. Phng Trnh ng Trn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204. Phng Trnh Elip. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Chuyn 4. Cc Bi Ton Lin Quan n Kho St Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231. Cc Tr Ca Hm S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232. Tng Giao Gia Hai Th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243. Tip Tuyn Ca Th Hm S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254. Bin Lun S Nghim Phng Trnh Bng Th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265. i Xng - Khong Cch & Cc Bi Ton Khc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Chuyn 5. Hm S Ly Tha. Hm S M & Hm S Lgarit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291. Ly Tha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292. Lgarit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303. Hm S Ly Tha. Hm S M & Hm S Lgarit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314. Phng Trnh & Bt Phng Trnh M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315. Phng Trnh & Bt Phng Trnh Lgarit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336. H Phng Trnh M & Lgarit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Chuyn 6. Phng Php Ta Trong Khng Gian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351. Ta Trong Khng Gian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352. Phng Trnh Mt Phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363. Phng Trnh ng Thng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384. Hnh Chiu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405. Gc V Khong Cch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Chuyn 7. Phng Trnh Lng Gic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451. Phng Trnh Lng Gic C Bn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452. Phng Trnh Lng Gic Thng Gp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463. Phng Trnh Lng Gic a V Phng Trnh Tch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474. Phng Trnh Lng Gic Cha n Mu. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485. Nghim Thuc Khong Cho Trc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

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Chuyn 8. Nguyn Hm - Tch Phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511. Nguyn Hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512. Mt S Phng Php Tm Nguyn Hm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523. Tch Phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524. Mt S Phng Php Tnh Tch Phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545. Tch Phn Ca Hm S Lng Gic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566. ng Dng Ca Tch Phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Chuyn 9. S Phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591. Dng i S Ca S Phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592. Phng Trnh Bc Hai Nghim Phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613. Dng Lng Gic Ca S Phc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Chuyn 10. Hnh Hc Khng Gian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 631. Quan H Song Song . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632. Quan H Vung Gc. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643. Th Tch Khi a Din. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654. Mt Nn - Mt Tr - Mt Cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Chuyn 11. T Hp - Xc Sut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691. Hon V - Chnh Hp - T Hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692. Xc Sut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703. Nh Thc Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Chuyn 12. Bt ng Thc & Gi Tr Ln Nht - Gi Tr Nh Nht. . . . . . . . . . . . . . . . . . . . . . 731. Bt ng Thc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732. Gi Tr Ln Nht - Gi Tr Nh Nht. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

PH LC 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

PH LC 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

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Chuyn 1

Kho St S Bin Thin V V ThHm S

1. Tnh n iu Ca Hm SA. Kin Thc Cn Nh

nh l 1.1. Cho hm s y = f(x) c o hm trn khong I. Nu f (x) > 0,x I th y = f(x) ng bin trn I. Nu f (x) < 0,x I th y = f(x) nghch bin trn I. Nu f (x) = 0,x I th y = f(x) khng i trn I.

Lu . Nu f (x) 0,x I v f (x) = 0 ti hu hn im ca I th y = f(x) ng bin trn I. Khong I trn c th c thay bi mt on hoc na khong vi gi thit b sung: Hm s y = f(x) lin

tc trn on hoc na khong .

B. K Nng C Bn

1. Tm cc khong n iu ca hm s. Tm tp xc nh. Tnh y. Tm cc im ti y bng 0 hoc khng xc nh. Lp bng bin thin. T bng bin thin rt ra kt lun.

2. iu kin hm s lun ng bin, nghch bin. Tm tp xc nh Df . Tnh y v ch ra y 0,x Df (hoc y 0,x Df ).

C. Bi Tp

1.1. Tm cc khong n iu ca cc hm s saua) y = 2x3 3x2 + 1. b) y = x3 3x+ 2. c) y = x3 + 3x2 + 3x.d) y = x4 2x2 + 3. e) y = x4 + 2x3 2x 1. f) y = x2 2x 3.g) y =

2x+ 3

x+ 2. h) y =

x+ 2

3x 1 . i) y =x2 4x+ 4

1 x .

1.2. Tm m hm s y = x3 + (m 1)x2 + (m2 4)x+ 9 lun ng bin trn R.1.3. Tm m hm s y = mx3 + (3m)x2 2x+ 2 lun nghch bin trn R.

1.4. Tm m hm s y =mx 2m x lun ng bin trn mi khong xc nh.

1.5. Tm m hm s y =mx 2x+m 3 lun nghch bin trn mi khong xc nh.

1.6. Tm m hm s y = x+ 2 +m

x 1 lun ng bin trn mi khong xc nh.

1.7. Tm m hm s y =mx+ 4

x+mnghch bin trn (; 1).

1.8. Tm m hm s y =mx 2x+m 3 nghch bin trn (1; +).

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Nguyn Minh Hiu

1.9. Tm a hm s y = x3 + 3x2 + ax+ a nghch bin trn on c di bng 1.

1.10. Tm m hm s y = x3 + 3x2 +mx+ 2 ng bin trn on c di bng 3.

2. Cc Tr Ca Hm SA. Kin Thc Cn Nh

nh l 1.2. Gi s hm s y = f(x) t cc tr ti x0. Khi , nu y = f(x) c o hm ti x0 th f (x0) = 0.

nh l 1.3. Gi s hm s y = f(x) lin tc trn khong (a; b) cha x0 v c o hm trn (a;x0), (x0; b). Khi Nu f (x) < 0,x (a;x0) v f (x) > 0,x (x0; b) th hm s y = f(x) t cc tiu ti x0. Nu f (x) > 0,x (a;x0) v f (x) < 0,x (x0; b) th hm s y = f(x) t cc i ti x0.

nh l 1.4. Gi s hm s y = f(x) c o hm cp mt trn (a; b) v c o hm cp hai khc 0 ti x0. Khi

Nu{f (x0) = 0f (x0) < 0

th hm s t cc i ti x0.

Nu{f (x0) = 0f (x0) > 0

th hm s t cc tiu ti x0.

Lu . Nu y(x0) = 0 th hm s c th t cc tr hoc khng t cc tr ti x0.

B. K Nng C Bn

1. Tm cc tr ca hm s. Tm tp xc nh. Tnh y. Tm cc im ti y bng 0 hoc khng xc nh. Lp bng bin thin. T bng bin thin rt ra kt lun.

2. iu kin hm s c cc tr, c k cc tr. S dng L 1.3 v L 1.4.

3. iu kin hm s t cc tr ti x0. Tnh y, y. Hm s t cc tr ti x0 y(x0) = 0 m. Thay m v x0 vo y kt lun.

Lu . Nu y(x0) = 0 th phi kim tra du ca y kt lun.

C. Bi Tp

1.11. Tm cc tr ca cc hm s saua) y = 2x3 3x2 + 1. b) y = x3 3x+ 2. c) y = x3 + 3x2 + 3x.d) y = x4 2x2 + 3. e) y = x4 + 2x3 2x 1. f) y = x2 2x 3.g) y =

2x+ 3

x+ 2. h) y =

x+ 2

3x 1 . i) y =x2 4x+ 4

1 x .

1.12. Tm m hm s y = x3 3mx2 + 3 (2m 1)x 2a) C cc tr. b) t cc tr ti x = 0. c) t cc i ti x = 1.

1.13. Cho hm s y =1

3x3 mx2 + (m2 m+ 1)x+ 1. Vi gi tr no ca m th hm s

a) t cc i ti x = 1. b) C cc i, cc tiu. c) Khng c cc tr.

1.14. Cho hm s y = x4 2 (m+ 1)x2 + 2m+ 1. Vi gi tr no ca m th hm sa) C ba im cc tr. b) t cc tiu ti x = 0. c) t cc tr ti x = 1.

1.15. Tm m hm s y = x4 + 2 (2m 1)x2 + 3 c ng mt cc tr.

1.16. (B-02) Tm m hm s y = mx4 +(m2 9)x2 + 10 c ba im cc tr.

1.17. Xc nh gi tr ca m hm s y =x2 +mx+ 1

x+ma) Khng c cc tr. b) t cc tiu ti x = 1. c) t cc i ti x = 2.

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

3. Gi Tr Ln Nht V Gi Tr Nh Nht Ca Hm SA. Kin Thc Cn Nh

nh ngha 1.5. Cho hm s y = f(x) xc nh trn tp hp D. Khi

M = maxxD

f(x){f(x) M,x Dx0 D : M = f(x0) . m = minxD f(x)

{f(x) m,x Dx0 D : m = f(x0) .

Lu . Mi hm s lin tc trn mt on u c gi tr ln nht v gi tr nh nht trn on . Trn khong hoc na khong hm s c th c hoc khng c gi tr ln nht v gi tr nh nht.

B. K Nng C Bn

1. Tm gi tr ln nht v gi tr nh nht ca hm s trn min D. Tnh y, y = 0 xi D. Lp bng bin thin. T bng bin thin rt ra kt lun.

2. Xt tnh n iu trn khong cho trc.PP1: Tnh y v ch ra y 0,x D (hoc y 0,x D). T y 0,x D m g(x),x D. Lp bng bin thin ca g(x) trn D. T bng bin thin rt ra kt lun.

PP2: Tnh y. Tm cc im ti y = 0 hoc khng xc nh. Lp bng bin thin. T bng bin thin rt ra kt lun.

Lu . m f(x),x D m max

xDf(x). m f(x),x D m min

xDf(x).

C. Bi Tp

1.18. Tm gi tr ln nht v gi tr nh nht (nu c) ca cc hm s sau:a) y = 1 + 8x 2x2 trn [1; 3]. b) y = x3 3x2 + 1 trn [2; 3]. c) y = 1 + 4x3 3x4 trn [2; 1].d) y = x3 3x2 + 1 trn (1; 4). e) y = x 5 + 1x trn (0; +). f) y = x 1x trn (0; 2].g) y =

4

1 + x2. h) y = x4 + 2x2 1. i) y = x+4 x2.

1.19. Tm gi tr ln nht v gi tr nh nht (nu c) ca cc hm s saua) y = x+

2 cosx trn

[0; pi2

]. b) y = 2 sinx 43 sin3x trn [0;pi]. c) y = sin4x 4sin2x+ 5.

d) y = sin4x+ cos4x. e) y = 5 sinx 12 cosx 5. f) y = sin2x+ sin 2x+ 2cos2x.

1.20. Cho parabol (P ) : y = x2 v im A (3; 0). Tm im M (P ) sao cho khong cch AM ngn nht v tnhkhong cch .

1.21. Tm m hm s y = x3 + 3x2 mx 4 ng bin trn (; 0).

1.22. (BT-79) Tm m hm s y = 13x3 + (m 1)x2 + (m 3)x 4 ng bin trn (0; 3).

1.23. Tm m hm s y = mx3 3 (m 1)x2 + 9 (m 2)x+ 1 ng bin trn [2; +).

1.24. Tm m hm s y = x3 + 3x2 + (m+ 1)x+ 4m ng bin trn (;2) v (2; +).

1.25. (BT-50) Tm m hm s y =mx2 + 6x 2

x+ 2nghch bin trn [1; +).

1.26. Tm m hm s y =x2 2mx+ 2m2 2

xm ng bin trn (1; +).

1.27. Tm a hm s y =x2 2ax+ 4a2

x 2a ng bin trn (2; +).

7

Nguyn Minh Hiu

4. ng Tim Cn Ca Th Hm SA. Kin Thc Cn Nh

nh ngha 1.6. ng thng y = y0 c gi l ng tim cn ngang ca th hm s y = f(x) nulim

x+ f(x) = y0 hoc limx f(x) = y0.

nh ngha 1.7. ng thng x = x0 c gi l ng tim cn ng ca th hm s y = f(x) nulimxx+0

f(x) = +; limxx+0

f(x) = ; limxx0

f(x) = + hoc limxx0

f(x) = .

nh ngha 1.8. ng thng y = ax+ b, (a 6= 0) c gi l ng tim cn xin ca th hm s y = f(x) nulim

x+ [f(x) (ax+ b)] = 0 hoc limx [f(x) (ax+ b)] = 0.

B. K Nng C Bn

1. Tm tim cn ngang v tim cn ng. Tm lim

x f(x)TCN. Tm limxx0f(x)TC.

Lu . x0 thng l mt nghim ca mu.

2. Tm tim cn xin.C1: Vit li hm s di dng y = ax+ b+ g(x). Ch ra lim

x [y (ax+ b)] = 0TCX.

C2: Tnh a = limx

f(x)

xv b = lim

x [f(x) ax]TCX.

C. Bi Tp

1.28. Tm tim cn (nu c) ca cc hm s sau

a) y =2x 1x 2 . b) y =

x 3x+ 2 . c) y =

3 4xx+ 1

.

d) y =x2 + x

x 1 . e) y =x+ 3

x+ 1. f) y = 2x 1 + 1

x.

g) y =x2 4x+ 4

1 x . h) y =x2 + x 1. i) y = x+

x2 + 2x.

1.29. Tm m th hm s y =mx2 2m (m 1)x 3m2 +m 2

x+ 2c tim cn xin i qua A (1;3).

1.30. Tm m hm s y =2x2 + (m+ 1)x 3

x+mc giao hai tim cn nm trn parabol (P ) : y = x2 + 2x 1.

1.31. (A-08) Tm m gc gia hai tim cn ca hm s y =mx2 +

(3m2 2)x 2x+ 3m

bng 450.

1.32. Tm m th hm s y =x2 +mx 1

x 1 c tim cn xin to vi cc trc to mt tam gic c din tchbng 4.

1.33. Tm m th hm s y =2x2 (5m 1)x+ 4m2 m 1

xm c tim cn xin to vi cc trc to mttam gic c din tch bng 4.

1.34. Cho hm s y =3x 1x 2 . Chng minh tch cc khong cch t im M nm trn th hm s n hai tim

cn khng i.

1.35. (A-07) Cho hm s y =x2 + 4x 3

x 2 . Chng minh tch cc khong cch t im M nm trn th hm sn hai tim cn l mt hng s.

1.36. Tm im M thuc th hm s y =3x 5x 2 tng khong cch t M n hai tim cn l nh nht.

1.37. Tm im M thuc th hm s y =x2 + 2x 2

x 1 tng khong cch t M n hai tim cn l nh nht.

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Chuyn 1. Kho St S Bin Thin V V Th Hm S

5. Kho St S Bin Thin V V Th Hm SA. Kin Thc Cn Nh

1. S kho st tng qut.1. Tp xc nh.2. S bin thin. Gii hn, tim cn (nu c). Bng bin thin (tnh o hm, lp bng bin thin, tnh n iu, cc tr).

3. th. Tng giao vi cc trc. Tnh i xng (nu c). im c bit (nu cn).

2. im un.

nh ngha 1.9. im U (x0; f(x0)) c gi l im un ca th hm s y = f(x) nu tn ti mt khong(a; b) cha im x0 sao cho trn mt trong hai khong (a;x0) v (x0; b) tip tuyn ca th ti im U nm phatrn th cn trn khong kia tip tuyn nm pha di th.

Mnh 1.10. Nu hm s y = f(x) c o hm cp hai trn mt khong cha x0, f (x0) = 0 v f (x) i dukhi qua im x0 th U (x0; f(x0)) l mt im un ca th hm s y = f(x).

B. Cc Dng Th Kho St

Hm s y = ax3 + bx2 + cx+ d (a 6= 0). Hm s y = ax4 + bx2 + c (a 6= 0).

O O

y y

x xU U

O O

y y

x x

Hm s y = ax+ bcx+ d

(c 6= 0, ad bc 6= 0). Hm s y = ax2 + bx+ c

dx+ e(a 6= 0, d 6= 0).

O O

y y

x x

I I

O O

y y

x x

I I

C. Bi Tp

1.38. Kho st s bin thin v v th ca cc hm s saua) y = x3 + 3x2 4. b) y = x3 + 3x 2. c) y = x3 + 1. d) y = x3 + 3x2 + 3x+ 1.e) y = x3 + x 2. f) y = 2x3 x 3. g) y = x3 + 3x2 1. h) y = 13x3 x2 3x 53 .

1.39. Kho st s bin thin v v th ca cc hm s saua) y = x4 2x2 3. b) y = x4 + 2x2 1. c) y = 12x4 + x2 32 . d) y = 3 2x2 x4.e) y = x4 + 2x2 2. f) y = 2x4 4x2 + 1. g) y = 2x4 4x2 + 1. h) y = x4 4x2 + 3.

1.40. Kho st s bin thin v v th ca cc hm s sau

a) y =4

2 x . b) y =x 32 x . c) y =

x+ 3

x 1 . d) y =x+ 22x+ 1

.

e) y =x 2x+ 1

. f) y =x+ 2

x 1 . g) y =2 xx+ 1

. h) y =x+ 3

x 2 .

9

Nguyn Minh Hiu

1.41. Kho st s bin thin v v th ca cc hm s sau

a) y =x2 + 2x+ 2

x+ 1. b) y =

x2 2x 3x 2 . c) y =

2x2 + 5x+ 4

x+ 2. d) y =

x2 2xx+ 1

.

e) y =x2 2xx 1 . f) y =

2x2 x+ 11 x . g) y = x+ 2 +

1

x 1 . h) y = x 1 +1

x+ 1.

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Chuyn 2

Phng Trnh - Bt Phng Trnh & HPhng Trnh i S

1. Phng Trnh & Bt Phng Trnh Khng Cha CnA. Phng Php Gii C Bn

1. a v phng trnh tch. Bin i a phng trnh v dng f(x).g(x) = 0. p dng cng thc f(x).g(x) = 0

[f(x) = 0g(x) = 0

.

2. t n ph. Chn n ph t = u(x) ph hp. a phng trnh v phng trnh theo n t bit cch gii (phng trnh c th vn cha x).

3. Phung php khong (i vi phng trnh cha n trong du gi tr tuyt i). Lp bng xt du cc biu thc trong du gi tr tuyt i. Xt phng trnh trn tng khong.

Lu . Nu phng trnh ch cha mt du tr tuyt i |f(x)| th xt hai trng hp f(x) 0 v f(x) < 0.

B. Bi Tp

2.1. Gii cc bt phng trnh saua) x2 6x+ 6 > 0. b) 4x2 + x 2 0.c) x4 4x3 + 3x2 + 8x 10 0. d) x4 + x2 + 4x 3 0.

2.2. Gii cc bt phng trnh sau

a)x 2

x2 9x+ 8 0. b)x2 3x 2

x 1 2x+ 2.

c)x+ 5

2x 1 +2x 1x+ 5

> 2. d)1

x2 5x+ 4 0.

2.13. Gii cc phng trnh saua) |9 x| = |6 5x|+ |4x+ 3|. b) x2 5x+ 4+ x2 5x = 4.c) |7 2x| = |5 3x|+ |x+ 2|. d) |x 1| 2 |x 2|+ 3 |x 3| = 4.e)x2 2x+ 1 +x2 + 4x+ 4 = 5. f)

x+ 2

x 1 +

x 2x 1 = 2.

2. Phng Trnh & Bt Phng Trnh Cha CnA. Phng Php Gii C Bn

1. S dng php bin i tng ng.

f(x) = g(x) { f(x) 0f(x) = g(x)

. f(x) = g(x) { g(x) 0f(x) = g2(x)

.

3f(x) = 3g(x) f(x) = g(x). 3f(x) = g(x) f(x) = g3(x). f(x) < g(x)

f(x) 0g(x) > 0f(x) < g2(x)

. f(x) > g(x){g(x) < 0f(x) 0{g(x) 0f(x) > g2(x)

.

2. t n ph Dng 1: t t = u(x), a phng trnh v n t (phng trnh c th vn cha n x). Dng 2. t u = u(x); v = v(x), a phng trnh v h theo n u v v.

3. S dng tnh n iu ca hm s. D on nghim (nu c). S dng tnh n iu ca hm s ch ra phng trnh ch c nghim d on (hoc ch ra PTVN).

4. nh gi hai v.

nh gi f(x) A; g(x) A. Khi f(x) = g(x){f(x) = Ag(x) = A

.

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Chuyn 2. Phng Trnh - Bt Phng Trnh & H Phng Trnh i S

B. Bi Tp

2.14. Gii cc phng trnh saua) xx 1 7 = 0. b) 2x+ 9 = 4 x+3x+ 1.c)

3x 35 x = 2x 4. d)

2x+

6x2 + 1 = x+ 1.e) 3

2x 1 + 3x 1 = 33x+ 1. f) 3x+ 1 + 3x+ 2 + 3x+ 3 = 0.2.15. Gii cc bt phng trnh sau

a)x2 4x 12 > 2x+ 3. b) x2 4x 12 x 4.

c) 3

6x 9x2 < 3x. d) x3 + 1 x+ 1.2.16. Gii cc bt phng trnh sau

a) (C-09)x+ 1 + 2

x 2 5x+ 1. b) (A-05) 5x 1x 1 > 2x 4.

c)

2x+

6x2 + 1 > x+ 1. d) (A-04)

2 (x2 16)x 3 +

x 3 > 7 x

x 3 .

2.17. Gii cc phng trnh saua) (D-05) 2

x+ 2 + 2

x+ 1x+ 1 = 4. b)

x 1 + 2x+ 2

x 1 2x+ 2 = 1.

c) x+x+ 12 +

x+ 14 = 9. d)

x+ 2

x 1 +

x 2x 1 = x+ 3

3.

2.18. Gii cc bt phng trnh sau

a)

x4 +x 4 8 x. b) (D-02) (x2 3x)2x2 3x 2 0.

c) (x 2)x2 + 4 < x2 4. d) (x+ 2)9 x2 x2 2x 8.e)x2 3x+ 2 +x2 4x+ 3 2x2 5x+ 4. f) x2 + x 2 +x2 + 2x 3 x2 + 4x 5.

2.19. Gii cc phng trnh saua) (D-06)

2x 1 + x2 3x+ 1 = 0. b)

7 x2 + xx+ 5 = 3 2x x2.

c)

2x2 + 8x+ 6 +x2 1 = 2x+ 2. d) 3

(2 +x 2) = 2x+x+ 6.

e) x2 + 3x+ 1 = (x+ 3)x2 + 1. f)

x2 7

x2+

x 7

x2= x.

2.20. Gii cc bt phng trnh sau

a)11 4x2

x< 3. b)

121 4x+ x2x+ 1

0.

c)2x

2x+ 1 1 > 2x+ 2. d)x2(

1 +

1 + x)2 > x 4.

2.21. Gii cc phng trnh saua) (x+ 5) (2 x) = 3x2 + 3x. b)

(x+ 1) (2 x) = 1 + 2x 2x2.

c)x+ 1 +

4 x+(x+ 1) (4 x) = 5. d) 3x 2 +x 1 = 4x 9 + 23x2 5x+ 2.

2.22. Gii cc phng trnh sau

a) x+

4 x2 = 2 + 3x4 x2. b) (x 3) (x+ 1) + 4 (x 3)

x+1x3 = 3.

c)4

x2+

x2

4 x2 +5

2

(4 x2x

+x

4 x2

)+ 2 = 0. d) (B-2011) 3

2 + x62 x+44 x2 = 103x.

2.23. Gii cc phng trnh saua) x2 + 3x+ 2 2x2 + 3x+ 5. b) x2 +2x2 + 4x+ 3 6 2x.c) x (x+ 1)x2 + x+ 4 + 2 0. d) x2 2x+ 8 6

(4 x) (2 + x) 0.

e)x

x+ 1 2x+ 1

x> 3. f)

x+ 2 +

x 1 + 2x2 + x 2 11 2x.

2.24. Gii cc phng trnh saua) x2 1 = 2xx2 2x. b) x2 1 = 2xx2 + 2x.c) (4x 1)x3 + 1 = 2x3 + 2x+ 1. d) x2 + 4x = (x+ 2)x2 2x+ 24.

2.25. Gii cc phng trnh saua) 3

2 x = 1x 1. b) (A-09) 2 33x 2 + 36 5x 8 = 0.c) 2

(x2 + 2

)= 5x3 + 1. d) 2

(x2 3x+ 2) = 3x3 + 8.

13

Nguyn Minh Hiu

2.26. Gii cc phng trnh saua) x2 +

x+ 5 = 5. b) x3 + 2 = 3 3

3x 2.

c) x3 + 1 = 2 3

2x 1. d) x 335 x3 (x+ 335 x3) = 30.2.27. Gii cc phng trnh, bt phng trnh sau

a) (B-2012) x+ 1 +x2 4x+ 1 3x. b) (A-2010) x

x

12 (x2 x+ 1) 1.c) 3x2 2 = 2 x3. d) x+

3 (1 x2) = 2 (1 2x2).

2.28. Gii cc phng trnh saua)

4x 1 +4x2 1 = 1. b) x 1 = x3 4x+ 5.c)

2x 1 +x2 + 3 = 4 x. d) x5 + x3 1 3x+ 4 = 0.e) x3 + 4x (2x+ 7)2x+ 3 = 0. f) (C-2012) 4x3 + x (x 1)2x+ 1 = 0.

2.29. Gii cc phng trnh saua)x2 2x+ 5 +x 1 = 2. b) x 2 +4 x = x2 6x+ 11.

c) 2(x 2 1)2 +x+ 6 +x 2 2 = 0. d) 5x3 + 3x2 + 3x 2 = 12x2 + 3x 12 .

3. H Phng Trnh i SA. Phng Php Gii C Bn

1. a v h mu mc. (H i xng loi I, h i xng loi II, h ng cp)2. Phng php th. Loi 1: Rt mt biu thc t mt phng trnh ri th vo phng trnh kia. Loi 2: Gii c th mt phng trnh ri th vo phng trnh kia. Loi 3. Th hng s.

3. t n ph.4. S dng tnh n iu ca hm s. Nu y = f(x) lun ng bin hoc nghch bin trn D th f(u) = f(v) u = v. Nu y = f(x) lun ng bin trn D cn y = g(x) lun nghch bin hoc khng i trn D th phng trnh

f(x) = g(x) c nhiu nht mt nghim trn D.

B. Bi Tp

2.30. Gii cc h phng trnh sau

a){x2 + y2 + xy = 7x+ y + xy = 5

. b){x+ y + xy = 1

x3 + y3 3(x y)2 + 2 = 0 .

c) (DB-05){x2 + y2 + x+ y = 4x (x+ y + 1) + y (y + 1) = 2

. d){x2 xy + y2 = 3 (x y)x2 + xy + y2 = 7(x y)2 .

2.31. Gii cc h phng trnh sau

a){x2 2y2 = 2x+ yy2 2x2 = 2y + x . b)

x 3y = 4y

x

y 3x = 4xy

.

c)

2x+ y =

3

x2

2y + x =3

y2

. d) (B-03)

3y =

y2 + 2

x2

3x =x2 + 2

y2

.

2.32. Gii cc h phng trnh sau

a){x2 xy = 22x2 + 4xy 2y2 = 14 . b)

{x2 2xy + 3y2 = 9x2 4xy + 5y2 = 5 .

c){x3 + y3 = 1x2y + 2xy2 + y3 = 2

. d) (DB-06){

(x y) (x2 + y2) = 13(x+ y)

(x2 y2) = 25 .

2.33. Gii cc h phng trnh sau

a){x+ y = 1x3 3x = y3 3y . b) (DB-06)

{x2 + 1 + y (y + x) = 4y(x2 + 1

)(y + x 2) = y .

c) (B-08){x4 + 2x3y + x2y2 = 2x+ 9x2 + 2xy = 6x+ 6

. d) (D-09){x (x+ y + 1) 3 = 0(x+ y)

2 5x2 + 1 = 0.

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Chuyn 2. Phng Trnh - Bt Phng Trnh & H Phng Trnh i S

2.34. Gii cc h phng trnh sau

a) (B-02){

3x y = x y

x+ y =x+ y + 2

. b) (A-03){x 1x = y 1y2y = x3 + 1

.

c){x2 + y2 + 2xyx+y = 1x+ y = x2 y . d)

{6x2 3xy + x+ y = 1x2 + y2 = 1

.

2.35. Gii cc h phng trnh sau

a) (DB-07){x4 x3y x2y2 = 1x3y x2 xy = 1 . b) (D-08)

{xy + x+ y = x2 2y2x

2y yx 1 = 2x 2y .

c) (D-2012){xy + x 2 = 02x3 x2y + x2 + y2 2xy y = 0 . d)

{x3 + 2y2 = x2y + 2xy

2x2 2y 1 + 3

y3 14 = x 2 .

2.36. Gii cc h phng trnh sau

a){x2 + y2 + xy = 1x3 + y3 = x+ 3y

. b){x3 + 2xy2 + 12y = 08y2 + x2 = 12

.

c) (DB-06){x3 8x = y3 + 2yx2 3 = 3 (y2 + 1) . d) (A-2011)

{5x2y 4xy2 + 3y3 2 (x+ y) = 0xy(x2 + y2

)+ 2 = (x+ y)

2 .

2.37. Gii cc h phng trnh sau

a) (B-09){xy + x+ 1 = 7yx2y2 + xy + 1 = 13y2

. b){

2x2 + x 1y = 2y y2x 2y2 = 2 .

c){

8x3y3 + 27 = 18y3

4x2y + 6x = y2. d)

{x3 y3 = 9x2 + 2y2 = x 4y .

2.38. Gii cc h phng trnh sau

a){x (3x+ 2y) (x+ 1) = 12x2 + 2y + 4x 8 = 0 . b)

{x+ y xy = 3x+ 1 +

y + 1 = 4

.

c) (C-2010){

2

2x+ y = 3 2x yx2 2xy y2 = 2 . d) (DB-05)

{ 2x+ y + 1x+ y = 1

3x+ 2y = 4.

e){x2 + y2 = 5y 1 (x+ y 1) = (y 2)x+ y . f) (A-08)

{x2 + y + x3y + xy2 + xy = 54x4 + y2 + xy (1 + 2x) = 54

.

2.39. Gii cc h phng trnh sau

a){

x+ 10 +y 1 = 11

x 1 +y + 10 = 11 . b){

x 1y = 8 x3(x 1)4 = y .

c) (A-2012){x3 3x2 9x+ 22 = y3 + 3y2 9yx2 + y2 x+ y = 12

. d) (A-2010){ (

4x2 + 1)x+ (y 3)5 2y = 0

4x2 + y2 + 2

3 4x = 7 .

4. Phng Trnh & H Phng Trnh Cha Tham SA. Kin Thc B Sung

Cho hm s y = f(x) lin tc trn D v c gi tr ln nht, gi tr nh nht trn D. Ta c: m = f(x) c nghim trn D min

xDf(x) m max

xDf(x).

m f(x) c nghim trn D m maxxD

f(x).

m f(x) c nghim trn D m minxD

f(x).

m f(x),x D m minxD

f(x).

m f(x),x D m maxxD

f(x).

B. Phng Php Gii C Bn

1. Phng php tam thc bc hai. Da vo nh l v du tam thc bc hai c iu kin ph hp cho tng bi ton.

2. Phng php chiu bin thin hm s. T bi ton bin i v rt m theo f(x). Lp BBT ca f(x). T BBT v cc kin thc b sung rt ra KL.

3. Phng php iu kin cn v . T tnh cht bi ton rt ra iu kin cn xy ra bi ton. Gii iu kin cn c m, thay li vo bi ton kim tra.

15

Nguyn Minh Hiu

C. Bi Tp

2.40. Tm m phng trnh(m5)x2 3mx+m+ 1 = 0.

a) C nghim. b) V nghim c) C hai nghim tri du.

2.41. Tm m phng trnh x2 + 2 (m+ 1)x+ 9m 5 = 0 c hai nghim m phn bit.2.42. Tm m phng trnh (m 2)x2 2mx+m+ 3 = 0 c hai nghim dng phn bit.2.43. Tm m phng trnh (m 2)x4 2 (m+ 1)x2 + 2m 1 = 0.

a) C mt nghim. b) C hai nghim phn bit. c) C bn nghim phn bit.

2.44. (D-04) Tm m h{

x+y = 1

xx+ y

y = 1 3m c nghim.

2.45. Tm m bt phng trnh

4x 2 +16 4x m c nghim.

2.46. Tm m phng trnh (x 3) (x+ 1) + 4 (x 3)

x+1x3 = m c nghim.

2.47. (DB-07) Tm m bt phng trnh m(x2 2x+ 2 + 1)+x (2 x) 0 c nghim thuc on [0; 1 +3].

2.48. (A-07) Tm m phng trnh 3x 1 +mx+ 1 = 2 4x2 1 c nghim thc.

2.49. (B-06) Tm m phng trnhx2 +mx+ 2 = 2x+ 1 c hai nghim thc phn bit.

2.50. (B-04) Tm m phng trnh m(

1 + x2 1 x2 + 2) = 21 x4 +1 + x2 1 x2 c nghim.2.51. (A-08) Tm m phng trnh 4

2x+

2x+ 2 4

6 x+ 26 x = m c hai nghim phn bit.

2.52. (DB-07) Tm m phng trnh 4x4 13x+m+ x 1 = 0 c ng mt nghim.

2.53. (B-07) Chng minh rng vi mi m > 0, phng trnh x2 + 2x 8 = m (x 2) c hai nghim phn bit.2.54. Chng minh rng vi mi m, phng trnh x4 + x3 2x2 + 3mxm2 = 0 lun c nghim.

2.55. (DB-04) Tm m h{x2 5x+ 4 03x2 mxx+ 16 = 0 c nghim.

2.56. (D-2011) Tm m h{

2x3 (y + 2)x2 + xy = mx2 + x y = 1 2m c nghim.

2.57. Tm m h

1 x2 + 2 31 x2 = m c nghim duy nht.

2.58. Tm m h{x = y2 y +my = x2 x+m c nghim duy nht.

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Chuyn 3

Phng Php Ta Trong Mt Phng

1. Ta Trong Mt PhngA. Kin Thc Cn Nh

Cho hai vect u (x1; y1) ,v (x2; y2) v ba im A (xA; yA) , B (xB ; yB) , C (xC ; yC). Ta c Hai vect bng nhau: u = v

{x1 = x2y1 = y2

.

Cc php ton vect: u v = (x1 x2; y1 y2); ku = (kx1; ky1). Hai vect cng phng: u ,v cng phng k 6= 0 : u = kv . Tch v hng ca hai vect: u .v = x1x2 + y1y2. Hai vect vung gc: uv u .v = 0. di vect: |u | =

x21 + y

21 .

Gc gia hai vect: cos (u ;v ) = u .v|u |.|v | . Ta vect: AB = (xB xA; yB yA). Khong cch gia hai im: AB =

AB = (xB xA)2 + (yB yA)2. Tnh cht trung im: I l trung im ca AB I

(xA + xB

2;yA + yB

2

).

Tnh cht trng tm: G l trng tm ABC G(xA + xB + xC

3;yA + yB + yC

3

).

B. Bi Tp

3.1. Trong mt phng Oxy, cho ba im A (1; 1) , B (2; 5) , C (4; 3). Tm ta im D sao cho AD = 3AB2AC.Tm ta im M sao cho

MA+ 2

MB = 5

MC.

3.2. Trong mt phng Oxy, cho ba im A (2; 5) , B (1; 1) , C (3; 3). Tm ta im D sao cho ABCD l hnh bnhhnh. Tm ta tm hnh bnh hnh .

3.3. Trong mt phng Oxy, cho hai im A (3; 2) , B (4; 3). Tm ta im M thuc trc Ox sao cho tam gicMAB vung ti M .

3.4. Trong mt phng Oxy, cho tam gic ABC c A (1;1) , B (5;3), nh C thuc trc Oy v trng tm G thuctrc Ox. Tm ta nh C v trng tm G.

3.5. Trong mt phng Oxy, cho A (1; 3) , B (0; 4) , C (3; 5) , D (8; 0). Chng minh ABCD l t gic ni tip.3.6. Trong mt phng Oxy, cho tam gic ABC c A (0; 6) , B (2; 0) , C (2; 0). Gi M l trung im AB, G l trngtm tam gic ACM v I l tm ng trn ngoi tip tam gic ABC. Chng minh GI vung gc vi CM .

3.7. (A-04) Trong mt phng Oxy, cho A (0; 2) , B(3;1). Tm to trc tm v tm ng trn ngoi tip

tam gic OAB.

3.8. (B-03) Trong mt phng Oxy, cho tam gic ABC vung cn ti A. Bit M (1;1) l trung im cnh BC vG(

23 ; 0)l trng tm tam gic. Tm to cc nh ca tam gic.

3.9. (D-04) Trong mt phng Oxy, cho tam gic ABC c A (1; 0) , B (4; 0) , C (0;m) ,m 6= 0. Tm to trngtm G. Tm m tam gic GAB vung ti G.

3.10. (D-2010) Trong mt phng Oxy, cho tam gic ABC c nh A (3;7), trc tm l H (3;1), tm ng trnngoi tip l I (2; 0). Xc nh ta nh C, bit C c honh dng.

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2. Phng Trnh ng ThngA. Kin Thc Cn Nh

1. Vect ch phng v php tuyn. Vect u 6= 0 c gi song song hoc trng vi gi l vect ch phng ca ng thng . Vect n 6= 0 c gi vung gc vi gi l vect php tuyn ca ng thng .

Lu . n (a; b) u (b;a) v ngc li.2. Phng trnh tham s ca ng thng.

ng thng qua M (x0; y0) v c vect ch phng u (a; b) c phng trnh tham s:{x = x0 + aty = y0 + bt

.

3. Phng trnh tng qut ca ng thng. Dng: ax+ by + c = 0 (a2 + b2 6= 0). Nhn xt: ng thng ax+ by + c = 0 c vect php tuyn n (a; b).

Cho x0 tu y0 ta c im M (x0; y0) thuc ng thng. ng thng qua M (x0; y0) v c VTPT n (a; b) c PT: a (x x0) + b (y y0) = 0. ng thng qua A (a; 0) v B (0; b) c phng trnh xa + yb = 1 gi l PT on chn. Trc Ox c phng trnh y = 0 v trc Oy c phng trnh x = 0.

4. Gc v khong cch.

Gc gia hai ng thng: cos (1; 2) = |n1.n2||n1| . |n2| .

Khong cch t mt im n mt ng thng: d (M,) = |ax0 + by0 + c|a2 + b2

.

Khong cch gia hai ng thng song song: d (1,2) = d (M,2), trong M l im bt k trn 1.

B. Bi Tp

3.11. Trong mt phng Oxy, cho ba im A (1; 2) , B (2; 3) v C (6; 2). Vit phng trnh ng thng qua A vsong song vi BC.

3.12. Trong mt phng Oxy, cho im A (3; 5). Vit phng trnh ng thng qua A ct hai tia Ox,Oy ln ltti M,N sao cho din tch tam gic OMN bng 30.

3.13. Trong mt phng Oxy, cho im A (8; 6). Lp phng trnh ng thng qua A v to vi hai trc ta mt tam gic c din tch bng 12.

3.14. (D-2010) Trong mt phng Oxy, cho im A (0; 2) v l ng thng i qua O. Gi H l hnh chiu vunggc ca A trn . Vit phng trnh ng thng , bit khong cch t H n trc honh bng AH.

3.15. (C-2011) Trong mt phng Oxy, cho ng thng d : x+ y + 3 = 0. Vit phng trnh ng thng i quaA (2;4) v to vi ng thng d mt gc bng 450.3.16. Trong mt phng Oxy, cho hai ng thng d1 : 2x y + 5 = 0; d2 : 3x+ 6y 1 = 0 v im M (2;1). Tmgiao im A ca d1, d2. Vit phng trnh ng thng qua M v ct d1, d2 ln lt ti B,C sao cho tam gicABC cn ti A.

3.17. Trong mt phng Oxy, cho tam gic ABC c nh B (4;5) v hai ng cao ln lt c phng trnh ld1 : 5x+ 3y 4 = 0 v d2 : 3x+ 8y + 13 = 0. Lp phng trnh cnh AC.3.18. Trong mt phng Oxy, cho tam gic ABC c phng trnh AB l 5x 3y + 2 = 0; cc ng cao qua nhA v B ln lt l d1 : 4x 3y + 1 = 0 v d2 : 7x+ 2y 22 = 0. Lp phng trnh hai cnh cn li.3.19. (C-2012) Trong mt phng Oxy, cho tam gic ABC. Cc ng thng BC,BB, BC ln lt c phngtrnh l y 2 = 0, x y+ 2 = 0, x 3y+ 2 = 0 vi B, C tng ng l chn cc ng cao k t B,C ca tam gicABC. Vit phng trnh cc ng thng AB,AC.

3.20. (D-09) Trong mt phng Oxy, cho tam gic ABC c M (2; 0) l trung im ca cnh AB. ng trung tuynv ng cao qua nh A ln lt c phng trnh l d1 : 7x 2y 3 = 0; d2 : 6x y 4 = 0. Vit phng trnhng thng AC.

3.21. Trong mt phng Oxy, cho tam gic ABC c nh A (1; 3) v hai trung tuyn k t B v C ln lt c phngtrnh d1 : x 2y + 1 = 0 v d2 : y 1 = 0. Lp phng trnh ng thng cha cnh BC.

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Chuyn 3. Phng Php Ta Trong Mt Phng

3.22. Trong mt phng Oxy, cho tam gic ABC c A (2;1) v hai ng phn gic trong ca gc B,C ln ltc phng trnh l d1 : x 2y + 1 = 0 v d2 : x+ y + 3 = 0. Lp phng trnh cnh BC.3.23. (D-2010) Trong mt phng Oxy, cho tam gic ABC vung ti A, c nh C (4; 1), phn gic trong gc Ac phng trnh x + y 5 = 0. Vit phng trnh ng thng BC, bit din tch tam gic bng 24 v nh A chong dng.

3.24. Trong mt phng Oxy, cho hnh bnh hnh c hai cnh l x+ 3y 6 = 0 v 2x 5y 1 = 0. Bit hnh bnhhnh c tm i xng I (3; 5), hy vit phng trnh hai cnh cn li ca hnh hnh.

3.25. (A-09) Trong mt phng Oxy, cho hnh ch nht ABCD c im I (6; 2) l giao im ca hai ng cho ACv BD. im M (1; 5) thuc ng thng AB v trung im E ca cnh CD thuc ng thng : x+ y 5 = 0.Vit phng trnh ng thng AB.

3.26. Trong mt phng Oxy, cho ng thng :{x = 2 2ty = 1 + 2t

v im M (3; 1). Tm im M (3; 1) sao cho

on MB l ngn nht.

3.27. (B-07) Trong mt phng Oxy, cho A (2; 2) v cc ng thng d1 : x+ y 2 = 0, d2 : x+ y 8 = 0. Tm imB d1 v C d2 sao cho tam gic ABC vung cn ti A.3.28. (B-04) Trong mt phng Oxy, cho A (1; 1) , B (4;3). Tm im C thuc d : x 2y 1 = 0 sao cho khongcch t C n ng thng AB bng 6.

3.29. (B-2011) Trong mt phng Oxy, cho cc ng thng : x y 4 = 0 v d : 2x y 2 = 0. Tm ta im N thuc d sao cho ng thng ON ct ng thng ti im M tha mn OM.ON = 8.

3.30. (A-06) Trong mt phng Oxy, cho ba ng thng d1 : x+ y + 3 = 0, d2 : x y 4 = 0, d3 : x 2y = 0. TmM thuc d3 sao cho khong cch t M n d1 bng hai ln khong cch t M n d2.

3.31. Trong mt phng Oxy, cho P (1; 6) , Q (3;4) v ng thng : 2x y 1 = 0. Tm to M trn saocho MP +MQ l nh nht. Tm to N trn sao cho |NP NQ| l ln nht.3.32. (C-09) Trong mt phng Oxy, cho tam gic ABC c C (1;2), ng trung tuyn k t A v ng caok t B ln lt c phng trnh l 5x+ y 9 = 0;x+ 3y 5 = 0. Tm ta cc nh A v B.3.33. (A-02) Trong mt phng Oxy, cho tam gic ABC vung ti A, ng thng cha BC c phng trnh

3x y 3 = 0, A v B thuc Ox, bn knh ng trn ni tip bng 2. Tm trng tm tam gic ABC.3.34. (B-2011) Trong mt phng Oxy, cho tam gic ABC c nh B

(12 ; 1). ng trn ni tip tam gic ABC

tip xc vi cc cnh BC,CA,AB tng ng ti cc im DEF . Cho D (3; 1) v ng thng EF c phng trnhy 3 = 0. Tm ta im A, bit A c tung dng.3.35. (B-08) Trong mt phng Oxy, tm to nh C ca tam gic ABC bit hnh chiu C ln ng thng ABl H(1;1), ng phn gic trong gc A l x y + 2 = 0 v ng cao k t B l 4x+ 3y 1 = 0.3.36. (D-2011) Trong mt phng Oxy, cho tam gic ABC c nh B (4; 1), trng tm G (1; 1) v ng thngcha phn gic trong ca gc A c phng trnh x y 1 = 0. Tm ta cc nh A v C.3.37. (A-2010) Trong mt phng Oxy, cho tam gic ABC cn ti A c nh A (6; 6); ng thng i qua trungim ca cc cnh AB v AC c phng trnh x+ y 4 = 0. Tm ta cc nh B v C, bit im E (1;3) nmtrn ng cao i qua nh C ca tam gic cho.

3.38. (B-09) Trong mt phng Oxy, cho tam gic ABC cn ti nh A c nh A (1; 4) v cc nh B,C thucng thng : x y 4 = 0. Xc nh to cc im B,C, bit din tch tam gic ABC bng 18.3.39. (B-02) Trong mt phng Oxy, cho hnh ch nht ABCD c tm I

(12 ; 0), AB : x2y+2 = 0, cnh AB = 2AD.

Tm to cc nh bit A c honh m.

3.40. (A-05) Trong mt phng Oxy, cho hai ng thng d1 : x y = 0, d2 : 2x + y 1 = 0. Tm cc nh hnhvung ABCD bit A thuc d1, B thuc d2 v B,D thuc trc honh.

3.41. (D-2012) Trong mt phng Oxy, cho hnh ch nht ABCD. Cc ng thng AC v AD ln lt c phngtrnh l x + 3y = 0 v x y + 4 = 0; ng thng BD i qua im M ( 13 ; 1). Tm ta cc nh ca hnh chnht ABCD.

3.42. (A-2012) Trong mt phng Oxy, cho hnh vung ABCD. Gi M l trung im ca cnh BC, N l im trncnh CD sao cho CN = 2ND. Gi s M

(112 ;

12

)v ng thng AN c phng trnh 2x y 3 = 0. Tm ta

im A.

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Nguyn Minh Hiu

3. Phng Trnh ng TrnA. Kin Thc Cn Nh

1. Phng trnh ng trn. Dng 1: (x a)2 + (y b)2 = R2 (R > 0) C tm I (a; b) v bn knh R =

R2.

Dng 2: x2 + y2 2ax 2by + c = 0 (a2 + b2 > c) C tm I (a; b) v bn knh R = a2 + b2 c.2. Tip tuyn vi ng trn. Tip tuyn ti M i qua im M v c vect php tuyn l IM . Phng trnh tip tuyn ti M l: (x0 a) (x x0) + (y0 b) (y y0) = 0.

3. Bn knh ng trn. im M thuc ng trn khi v ch khi R = IM . ng thng tip xc vi ng trn khi v ch khi R = d (I; ).

B. Bi Tp

3.43. (B-06) Trong mt phng Oxy, cho ng trn (C) : x2 + y2 2x 6y + 6 = 0 v im M (3; 1). Gi T1, T2l cc tip im v t M n (C). Lp phng trnh ng thng T1T2.

3.44. (D-2011) Trong mt phng Oxy, cho im A (1; 0) v ng trn (C) : x2 + y2 2x+ 4y 5 = 0. Vit phngtrnh ng thng ct (C) ti hai im M,N sao cho tam gic AMN vung cn ti A.

3.45. (C-2012) Trong mt phng Oxy, cho ng trn (C) : x2 + y2 2x 4y + 1 = 0 v ng thng d :4x 3y +m = 0. Tm m d ct (C) ti hai im A,B sao cho AIB = 1200, vi I l tm ca (C).3.46. (D-09) Trong mt phng Oxy, cho ng trn (C) : (x 1)2 + y2 = 1. Gi I l tm ca (C). Xc nh to im M (C) sao cho IMO = 300.3.47. (A-09) Trong mt phngOxy, cho ng trn (C) : x2+y2+4x+4y+6 = 0 v ng thng : x+my2m+3 =0, vi m l tham s thc. Gi I l tm ca ng trn (C). Tm M ct (C) ti hai im phn bit A,B saocho din tch tam gic IAB ln nht.

3.48. (A-2011) Trong mt phng Oxy, cho ng thng : x+y+ 2 = 0 v ng trn (C) : x2 +y24x2y = 0.Gi I l tm ca (C), M l im thuc . Qua M k cc tip tuyn MA v MB n (C), (A,B l tip im). Tmta im M , bit t gic MAIB c din tch bng 10.

3.49. (B-09) Trong mt phng Oxy, cho ng trn (C) : (x 2)2 + y2 = 45 v hai ng thng 1 : x y = 0,2 : x 7y = 0. Xc nh to tm K v tnh bn knh ca ng trn (C1), bit ng trn (C1) tip xc vihai ng thng 1,2 v tm K thuc ng trn (C).

3.50. (A-07) Trong mt phng Oxy, cho tam gic ABC c A (0; 2) , B (2;2) , C (4;2). Gi H l chn ng caov t B v M,N l trung im AB,BC. Vit phng trnh ng trn qua H,M,N .

3.51. (D-2012) Trong mt phng Oxy, cho ng thng d : 2x y + 3 = 0. Vit phng trnh ng trn c tmthuc d, ct trc Ox ti A v B, ct trc Oy ti C v D sao cho AB = CD = 2.

3.52. (B-2012) Trong mt phng Oxy, cho cc ng trn (C1) : x2 + y2 = 4, (C2) : x2 + y2 12x + 18 = 0 vng thng d : x y 4 = 0. Vit phng trnh ng trn c tm thuc (C2) tip xc vi d v ct (C1) ti haiim phn bit A v B sao cho AB vung gc vi d.

3.53. (A-2010) Trong mt phng Oxy, cho hai ng thng d1 :

3x+ y = 0 v d2 :

3x y = 0. Gi (T ) l ngtrn tip xc vi d1 ti A, ct d2 ti hai im B,C sao cho tam gic ABC vung ti B. Vit phng trnh ca (T ),bit tam gic ABC c din tch bng

3

2 v im A c honh dng.

4. Phng Trnh ElipA. Kin Thc Cn Nh

O

y

xF1 F2A1 A2

B1

B2

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Chuyn 3. Phng Php Ta Trong Mt Phng

Phng trnh chnh tc ca elip: x2

a2+y2

b2= 1

(b2 = a2 c2).

Trong :Cc nh: A1(a; 0), A2(a; 0), B1(0;b), B2(0; b).Cc tiu im: F1(c; 0), F2(c; 0).Trc ln: A1A2 = 2a.Trc nh: B1B2 = 2b.Tiu c: F1F2 = 2c.Tm sai: e =

c

a.

Bn knh qua tiu: MF1 = a+cx

a,MF2 = a cx

a.

B. Bi Tp

3.54. Tm ta cc tiu im, cc nh, di trc ln, di trc b ca mi elip c phng trnh sau

a)x2

25+y2

4= 1. b)

x2

9+y2

4= 1. c) x2 + 4y2 = 4.

3.55. Vit phng trnh chnh tc ca cc ng elip (E) trong mi trng hp saua) (E) c di trc ln bng 8 v tm sai e =

3

2 .b) (E) c di trc b bng 8 v tiu c bng 4.c) (E) c mt tiu im l F

(3; 0)v i qua im M

(1;

32

).

3.56. (A-08) Trong mt phng Oxy, lp phng trnh chnh tc ca elip c tm sai bng

53 v hnh ch nht c

s c chu vi 20.

3.57. (D-05) Trong mt phng Oxy, cho im C(2; 0) v elip (E) :x2

4+y2

1= 1. Tm A,B thuc (E) bit A,B i

xng nhau qua trc honh v tam gic ABC u.

3.58. (B-2010) Trong mt phng Oxy, cho im A(2;

3)v elip (E) :

x2

3+y2

2= 1. Gi F1 v F2 l cc tiu im

ca (E) (F1 c honh m); M l giao im c tung dng ca ng thng AF1 vi (T ); N l im i xngca F2 qua M . Vit phng trnh ng trn ngoi tip tam gic ANF2.

3.59. (B-2012) Trong mt phng Oxy, cho hnh thoi ABCD c AC = 2BD v ng trn tip xc vi cc cnhca hnh thoi c phng trnh x2 + y2 = 4. Vit phng trnh chnh tc ca elip (E) i qua cc nh A,B,C,D cahnh thoi. Bit A thuc Ox.

3.60. (A-2011) Trong mt phng Oxy, cho (E) :x2

4+y2

1= 1. Tm ta cc im A v B thuc (E), c honh

dng sao cho tam gic OAB cn ti O v c din tch ln nht.

3.61. (A-2012) Trong mt phng Oxy, cho ng trn (C) : x2 + y2 = 8. Vit phng trnh chnh tc ca elip (E),bit rng (E) c di trc ln bng 8 v (E) ct (C) ti bn im to thnh bn nh ca mt hnh vung.

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Nguyn Minh Hiu

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Chuyn 4

Cc Bi Ton Lin Quan n Kho StHm S

1. Cc Tr Ca Hm SA. Kin Thc B Sung

Cch tnh tung cc tr: Nu y = f (x).g(x) + r(x) th y0 = r(x0). Nu y = u(x)

v(x)th y0 =

u(x0)v(x0)

.

B. Bi Tp

4.1. Tm m hm s y = x3 3 (m+ 1)x2 + 9xm t cc tr ti x1, x2 tha |x1 x2| 2.4.2. Tm m hm s y = x3 +2 (m 1)x2 +(m2 4m+ 1)x+1 t cc tr ti x1, x2 tha 1x1 + 1x2 = 12 (x1 + x2).4.3. (D-2012) Tm m hm s y = 23x

3 mx2 2 (3m2 1)x + 23 c hai im cc tr x1 v x2 sao cho x1x2 +2 (x1 + x2) = 1.

4.4. Tm m hm s y = x3 + (2m+ 1)x2 (m2 3m+ 2)x 4 c hai cc tr nm v hai pha trc tung.4.5. (DB-05) Tm m hm s y =

x2 + 2mx+ 1 3m2xm c hai cc tr nm v hai pha trc tung.

4.6. (DB-04) Tm m hm s y = x3 3 (m+ 1)x2 + 3m (m+ 2)x+ 1 t cc i, cc tiu ti cc im c honh dng.

4.7. Tm m hm s y =mx2 + 3mx+ 2m+ 1

x 1 c cc i, cc tiu nm v hai pha ca trc Ox.

4.8. (A-02) Cho hm s y = x3 + 3mx2 + 3 (1m2)x+m3m2. Vit phng trnh ng thng i qua hai imcc tr ca hm s.

4.9. Tm m hm s y = x3 32mx2 + 12m3 c cc i, cc tiu i xng nhau qua ng thng y = x.4.10. (B-07) Tm m hm s y = x3 + 3x2 + 3 (m2 1)x 3m2 1 c cc i, cc tiu v cc im cc tr cchu gc to .

4.11. Tm m hm s y = x3 3mx 3m+ 1 c cc tr ng thi chng cch u ng thng d : x y = 0.4.12. (D-2011) Tm m hm s y = x4 2 (m+ 1)x2 +m c ba cc tr A,B,C sao cho OA = BC, trong O lgc ta v A thuc trc tung.

4.13. Tm m hm s y = x4 2mx2 + 2m+m4 c cc i, cc tiu lp thnh tam gic u.4.14. (A-2012) Tm m hm s y = x4 2 (m+ 1)x2 + m2 c ba im cc tr to thnh ba nh ca mt tamgic vung.

4.15. (A-07) Tm m hm s y =x2 + 2 (m+ 1)x+m2 + 4m

x+ 2c cc i cc tiu ng thi cc im cc tr cng

vi gc to to thnh mt tam gic vung.

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Nguyn Minh Hiu

4.16. (B-2012) Tm m hm s y = x3 3mx2 + 3m3 c hai im cc tr A v B sao cho tam gic OAB c dintch bng 48.

4.17. (A-05) Tm m hm s y = mx + 1x c cc i cc tiu ng thi khong cch t im cc tiu n timcn xin bng 1

2.

4.18. (B-05) Chng minh rng vi mi m bt k, hm s y =x2 + (m+ 1)x+m+ 1

x+ 1lun c im cc i, im

cc tiu v khong cch gia hai im bng

20.

4.19. Tm m hm s y = 13x3 mx2 x+m+ 1 c khong cch gia cc i, cc tiu l nh nht.

2. Tng Giao Gia Hai ThA. Kin Thc Cn Nh

1. Giao im ca hai th. Honh giao im ca (C1) : y = f(x) v (C2) : y = g(x) l nghim ca phng trnh f(x) = g(x). S giao im ca (C1) v (C2) bng s nghim ca phng trnh f(x) = g(x).

Lu . Phng trnh f(x) = g(x) gi l phng trnh honh giao im.

2. S tip xc gia hai th.

th hai hm s y = f(x) v y = g(x) tip xc nhau ti M (x0; y0){f(x0) = g(x0)f (x0) = g(x0)

.

B. Bi Tp

4.20. Tm giao im ca th hm s y = x3 + 3x2 3x 2 v parabol y = x2 4x+ 2.4.21. Tm m th hm s y = mx3 x2 2x+ 8m ct trc honh ti ba im phn bit.4.22. Tm m th hm s y = x3 3mx2 1 ct trc Ox ti ba im phn bit.4.23. Tm a th hm s y = x3 + ax+ 3 ct ng thng y = 1 ti ng mt im.

4.24. (D-06) Gi d l ng thng i qua A (3; 20) v c h s gc m. Tm m d ct th hm s y = x33x+ 2ti ba im phn bit.

4.25. (A-2010) Tm m hm s y = x3 2x2 + (1m)x+m c th ct trc honh ti ba im phn bit chonh x1, x2, x3 tho mn iu kin x21 + x

22 + x

23 < 4.

4.26. Tm m th hm s y = x3 mx2 + 4x+ 4m 16 ct Ox ti ba im phn bit c honh ln hn 1.

4.27. Chng minh rng th hm s y =x 1x+ 1

lun ct ng thng y = m x vi mi gi tr ca m.

4.28. Tm m ng thng qua A (2; 2) v c h s gc m ct th hm s y = 2x 1x+ 1

ti hai im thuc hai

nhnh phn bit.

4.29. (D-2011) Tm k ng thng y = kx + 2k + 1 ct th hm s y =2x+ 1

x+ 1ti hai im phn bit A,B

sao cho khong cch t A v B n trc honh bng nhau.

4.30. Chng minh vi mi gi tr ca m th ng thng y = 2x+m lun ct th hm s y =x+ 3

x+ 1ti hai im

phn bit M,N . Xc nh m sao cho di MN l nh nht.

4.31. (B-09) Tm m ng thng y = x+m ct th hm s y = x2 1x

ti hai im phn bit A,B sao choAB = 4.

4.32. Tm m ng thng y = x+m ct th hm s y = x2 2x+ 2x 1 ti hai im A,B i xng nhau qua

ng thng y = x+ 3.

4.33. Tm m th hm s y = x4 + 2mx2 2m+ 1 ct trc honh ti hai im phn bit.

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Chuyn 4. Cc Bi Ton Lin Quan n Kho St Hm S

4.34. Tm m th hm s y = x4 (3m+ 4)x2 +m2 ct trc honh ti bn im phn bit c honh lpthnh cp s cng.

4.35. (D-09) Tm m ng thng y = 1 ct th hm s x4 (3m+ 2)x2 + 3m ti bn im phn bit chonh nh hn 2.

4.36. (DB-08) Tm m th hm s y = x4 8x2 + 7 tip xc vi ng thng y = mx 9.

4.37. (D-02) Tm m th hm s y =(2m 1)xm2

x 1 tip xc vi ng thng y = x.

4.38. Tm m th hm s y = x4 2mx2 +m3 m2 tip xc vi trc honh ti hai im phn bit.

3. Tip Tuyn Ca Th Hm SA. Kin Thc Cn Nh

H s gc ca tip tuyn ti M (x0; y0) l k = y(x0). Phng trnh tip tuyn ti M (x0; y0) l y = y(x0) (x x0) + y0.

B. Bi Tp

4.39. (B-04) Vit phng trnh tip tuyn ca th hm s y = 13x3 2x2 + 3x (C) ti tm i xng v chng

minh rng l tip tuyn ca (C) c h s gc nh nht.

4.40. (DB-08) Cho hm s y = x3 + 3mx2 + (m+ 1)x + 1. Tm m tip tuyn ti im c honh x = 1 iqua im A (1; 2).

4.41. (TN-08) Vit phng trnh tip tuyn ca th hm s y =3x 2x+ 1

ti im c tung bng 2.

4.42. (DB-06) Cho hm s y =x+ 3

x+ 1. Tip tuyn ti im (S) bt k ca (C) ct hai tim cn ca (C) ti P v

Q. Chng minh S l trung im PQ.

4.43. Cho hm s (Cm) : y = x3 + 1m (x+ 1). Vit phng trnh tip tuyn ca th (Cm) ti giao im ca(Cm) vi Oy. Tm m tip tuyn ni trn chn hai trc to to thnh tam gic c din tch bng 8.

4.44. (TN-09) Vit phng trnh tip tuyn ca th hm s y =2x+ 1

x 2 , bit h s gc ca tip tuyn bng 5.

4.45. Vit phng trnh tip tuyn ca th hm s y =x+ 32x 1 bit tip tuyn song song vi ng phn gic

gc phn t th hai ca mt phng to .

4.46. (C-2012) Vit phng trnh tip tuyn d ca th hm s y =2x+ 3

x+ 1, bit d vung gc vi ng thng

y = x+ 2.

4.47. (D-05) Cho hm s y = 13x3 m2 x2 + 13 c th (Cm). Gi M l im thuc (Cm) c honh bng 1.

Tm m tip tuyn ca (Cm) ti im M song song vi ng thng 5x y = 0.

4.48. (B-06) Vit phng trnh tip tuyn ca th hm s y =x2 + x 1x+ 2

(C). Bit rng tip tuyn vung

gc vi tim cn xin ca (C).

4.49. (DB-07) Lp phng trnh tip tuyn ca th hm s y =x

x 1 sao cho tip tuyn v hai tim cn ctnhau to thnh mt tam gic cn.

4.50. Tm m (Cm) : y = x3 + 3x2 +mx+ 1 ct ng thng y = 1 ti ba im phn bit C (0; 1) , D,E sao chocc tip tuyn vi (Cm) ti D v E vung gc vi nhau.

4.51. (A-2011) Cho hm s (C) : y =x+ 12x 1 . Chng minh rng vi mi m ng thng y = x+m lun ct th

(C) ti hai im phn bit A v B. Gi k1, k2 ln lt l h s gc ca cc tip tuyn vi (C) ti A v B. Tm m tng k1 + k2 t gi tr ln nht.

4.52. (B-08) Vit phng trnh tip tuyn ca th hm s y = 4x3 6x2 + 1, bit tip tuyn qua M (1;9).

25

Nguyn Minh Hiu

4.53. (DB-07) Lp phng trnh tip tuyn ca th hm s y =x+ 12x+ 1

, bit tip tuyn qua giao im ca tim

cn ng v trc Ox.

4.54. (DB-05) Cho hm s y =x2 + 2x+ 2

x+ 1c th (C). Gi I l giao hai tim cn. Chng minh rng khng c

tip tuyn no ca (C) i qua I.

4.55. Tm trn ng thng y = 4 cc im k c ba tip tuyn n (C) : y = x3 12x+ 12.

4. Bin Lun S Nghim Phng Trnh Bng ThA. K Nng C Bn

1. V th hm s y = f (|x|). V th hm s y = f(x) v b phn th bn tri Oy. i xng phn th bn phi Oy qua Oy.

2. V th hm s y = |f(x)|. V th hm s y = f(x). i xng phn th di Ox qua Ox v b phn th di Ox.

3. Da vo th bin lun s nghim ca phng trnh f(x) = k(m). V th hm s y = f(x) v ng thng y = k(m) song song vi Ox. S nghim phng trnh f(x) = k(m) l s giao im ca th y = f(x) vi ng thng y = k(m). Da vo mi tng quan trong hnh v bin lun.

B. Bi Tp

4.56. Kho st s bin thin v v th hm s y = x3 3x2 1. Bin lun theo k s nghim ca phng trnhx3 3x2 k = 0.4.57. Kho st s bin thin v v th hm s y = 2x3 3x2 + 1. Bin lun theo m s nghim phng trnh4x3 6x2 m = 0.4.58. Kho st s bin thin v v th hm s y = x4 + 2x2 + 3. Bin lun theo m s nghim ca phng trnhx4 2x2 +m 1 = 0.4.59. Kho st s bin thin v v th hm s y = x4 4x2 + 3. Tm m phng trnh 12x4 2x2 +m = 0 cbn nghim phn bit.

4.60. (DB-06) Kho st s bin thin v v th hm s y =x2 + 2x+ 5

x+ 1. Tm m phng trnh sau c hai

nghim dng phn bit x2 + 2x+ 5 =(m2 + 2m+ 5

)(x+ 1).

4.61. (A-06) Kho st s bin thin v v th hm s y = 2x3 9x2 + 12x 4. Tm m phng trnh sau csu nghim phn bit 2|x|3 9x2 + 12 |x| = m.4.62. Kho st s bin thin v v th hm s y = 2x3+3x22. Tmm phng trnh 2|x|33x2+2 (m+ 1) = 0c ng bn nghim.

4.63. (DB-03) Kho st s bin thin v v th hm s y =2x2 4x 3

2 (x 1) . Tm m phng trnh 2x2 4x

3 + 2m |x 1| = 0 c hai nghim phn bit.

4.64. Kho st s bin thin v v th hm s y =x2 + 3x+ 3

x+ 1. Tm m phng trnh x

2+3x+3|x+1| = m c bn

nghim phn bit.

4.65. Kho st s bin thin v v th hm s y = x3 3x2 + 4. Tm m phng trnh sau c bn nghim phnbit |x 1|3 3 |x 1| m = 0.4.66. Kho st s bin thin v v th hm s y = x33x+1. Tmm phng trnh x3 3x+ 12m2+m = 0c ba nghim phn bit.

4.67. (B-09) Kho st s bin thin v v th hm s y = 2x4 4x2. Vi cc gi tr no ca m, phng trnhx2x2 2 = m c ng su nghim thc phn bit.

4.68. Kho st s bin thin v v th hm s y = x4 4x2 + 3. Tm m phng trnh x4 4x3 + 3 = m cng tm nghim.

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Chuyn 4. Cc Bi Ton Lin Quan n Kho St Hm S

5. i Xng - Khong Cch & Cc Bi Ton Khc4.69. Tm m hm s y =

m2x 2x 1 qua im A (2; 6).

4.70. Tm cc h s m,n sao cho hm s y = x3 +mx+ n t cc tiu ti im x = 1 v th ca n i quaim M (1; 4).

4.71. Chng minh rng im un ca th hm s (C) : y = x3 6x2 + 9x l tm i xng ca n.

4.72. Chng minh rng th hm s (C) : y =2x+ 1

x+ 1nhn giao im I ca hai tim cn lm tm i xng.

4.73. (D-04) Tm m tm i xng ca th hm s y = x3 3mx2 + 9x+ 1 thuc ng thng y = x+ 1.

4.74. Tm m th hm s y = x3

m+ 3x2 2 nhn I (1; 0) lm im un.

4.75. Tm cc im trn th hm s y =2x 1x 1 c ta l cc s nguyn.

4.76. Tm trn th hm s y =x2 + 3x 1

x 1 cc im c to nguyn.

4.77. Tm im c nh ca h ng cong (Cm) : y = x3 + 2 (m 1)x2 + (m2 4m+ 1)x 2 (m2 + 1).4.78. Chng minh rng vi mi m 6= 1, h ng cong (Cm) : y = mx 1

xm lun i qua hai im c nh. Gi Ml giao im ca hai tim cn ca (Cm), tm tp hp im M khi m thay i.

4.79. Cho h ng cong (Cm) : y = mx3 + (1m)x. Tm cc im trn mt phng ta sao cho khng cng no ca (Cm) i qua.

4.80. (DB-06) Tm trn th hm s y = 13x3 + x2 + 3x 11

3hai im phn bit M,N i xng nhau qua Oy.

4.81. Tm trn th hm s y = x3 + 3x 2 hai im i xng nhau qua M (2; 18).

4.82. Tm trn th hm s y =3x+ 1

x 2 hai im i xng nhau qua M (2;1).

4.83. Cho hm s y =x+ 1

x 1 c th (C). Tm trn (C) hai im phn bit A,B i xng nhau qua ng thngd : x+ 2y 3 = 0.4.84. (B-03) Tm m th hm s y = x3 3x2 +m c hai im phn bit i xng nhau qua gc to .

4.85. (DB-04) Tm trn th hm s y =x

x+ 1nhng im M sao cho khong cch t M n ng thng

d : 3x+ 4y = 0 bng 1.

4.86. Cho hm s y =4x+ 1

x+ 1c th (C). Tm trn (C) cc im cch u hai trc ta .

4.87. Cho hm s y =x2 x+ 1x 1 . Tm im M trn th hm s sao cho khong cch t M n giao im I ca

hai tim cn l nh nht.

4.88. Cho hm s y =3x 5x 2 c th (C). Tm im M trn (C) tng khong cch t M n hai tim cn l

nh nht.

4.89. Cho hm s y =x 1x+ 1

c th (C). Tm im M trn (C) tng khong cch t M n hai trc to

l nh nht.

4.90. Tm hai im trn hai nhnh th hm s y =x 2x 1 c khong cch b nht.

27

Nguyn Minh Hiu

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Chuyn 5

Hm S Ly Tha. Hm S M & HmS Lgarit

1. Ly ThaA. Kin Thc Cn Nh

1. Cc nh ngha. Ly tha vi s m nguyn dng: an = a.a...a

n tha s

(a R, n N).

Ly tha vi s m 0: a0 = 1 (a 6= 0). Ly tha vi s m nguyn m: an = 1an (a 6= 0, n N). Cn bc n: b l cn bc n ca a bn = a.Lu : Khi n l th a c ng mt cn bc n l n

a.

Khi n chn th a < 0 khng c cn bc n.a = 0 c mt cn bc n l 0.a > 0 c hai cn bc n l na.

Ly tha vi s m hu t: amn = nam (a > 0;m,n Z;n 2). Ly tha vi s m thc: a = lim

n+ arn

(a > 0; (rn) Q; lim

n+ rn = ).

2. Cc tnh cht ca ly tha vi s m thc.Cho hai s a, b > 0 v , l nhng s thc tu . Ta c

a.a = a+ . a

a= a .

(a) = a . (ab) = a.b.

(ab

)=a

b.

Nu a > 1 th a > a > . Nu 0 < a < 1 th a > a < . Nu > 0 th 0 < a < b a < b. Nu < 0 th 0 < a < b a > b.

B. Bi Tp

5.1. Tnh gi tr cc lu tha sau

a) (0, 04)1,5 (0, 125) 23 . b)(

1

16

)0,75+

(1

8

) 43.

c) 2723 +

(1

16

)0,75 250,5. d) (0, 5)4 6250,25

(2

1

4

)1 12.

e) 810,75 +(

1

125

) 13(

1

32

) 35. f)

102+

7

22+

7.51+

7.

g)(

42

3 4

31).22

3. h)

(6

25 + 4

6 3

1 + 2

6

)3

1 2

6.

5.2. Rt gn cc biu thc sau

a)x54 y + xy

54

4x+ 4y. b)

a13

b+ b

13a

6a+ 6b

.

29

Nguyn Minh Hiu

c)ab

4a 4b

a 4ab4a+ 4b. d)

a b3a 3b

a+ b3a+ 3b.

e)

(a2

3 1)(

a2

3 + a

3 + a3

3)

a4

3 a3 .f)(

a+ b3a+ 3b 3ab

):(

3a 3b)2

.

g)a 1a34 + a

12

.

a+ 4a

a+ 1.a

14 + 1. h)

(a+

b32

a12

)(a12 b 12a12

+b12

a12 b 12

) 23.

5.3. Hy so snh cc cp s saua) 3

10 v 5

20. b) 4

13 v 5

23.c) 3600 v 5400. d) 3

7 +

15 v

10 + 3

28.

5.4. Tnh A =a+ b+ c+ 2

ab+ bc+

a+ b+ c 2ab+ bc, (a, b, c > 0, a+ c > b)

2. LgaritA. Kin Thc Cn Nh

1. nh ngha. = logab a = b (a, b > 0; a 6= 1).2. Tnh cht. loga1 = 0. logaa = 1. alogab = b. loga (a) = . Khi a > 1 th logab > logac b > c. Khi 0 < a < 1 th logab > logac b < c.

3. Quy tc tnh. loga (bc) = logab+ logac. loga bc = logab logac. loga 1b = logab. logab = logab. loga n

b = 1n logab. logab = logac.logcb.

logab = 1logba . logab = 1 logab.4. Lgarit thp phn v lgarit t nhin. Lgarit thp phn: L lgarit c c s a = 10. K hiu: log x hoc lg x. Lgarit t nhin: L lgarit c c s a = e. K hiu: lnx.

B. Bi Tp

5.5. Tnha) log3

4

3. b) 2log27 log 1000. c) log258.log85.d) log 45 2 log 3. e) 3log2log416 + log 12 2. f) log248 13 log227.g) 5 ln e1 + 4 ln

(e2e). h) log 72 2 log 27256 + log

108. i) log 0, 375 2 log0, 5625.

5.6. n gin biu thc

a)log24 + log2

10

log220 + log28. b)

log224 12 log272log318 13 log372

. c)(

log72 +1

log57

)log 7.

d) loga

(a2. 3a.

5a4

4a

). e) log5log5

5

5

...

5

5 n du cn

. f) 92log34+4log812.

g) 161+log45 + 412 log23+3log55. h)

(81

14 12 log94 + 25log1258

)49log72. i) 72

(49

12 log79log76 + 5log54

).

5.7. So snh cc cp s sau:a) log3

65 v log3

56 . b) log 12 e v log 12pi. c) log210 v log530.

d) log53 v log0,32. e) log35 v log74. f) log310 v log857.

5.8. Tnh log41250 theo a, bit a = log25.

5.9. Tnh log54168 theo a, b, bit a = log712, b = log1224.

5.10. Tnh log14063 theo a, b, c, bit a = log23, b = log35, c = log72.

5.11. Tnh log 325135 theo a, b, bit a = log475, b = log845.

5.12. Chng minh rng ab+ 5 (a b) = 1, bit a = log1218, b = log2454.5.13. Cho y = 10

11log x , z = 10

11log y . Chng minh rng x = 10

11log z .

5.14. Cho a, b, c > 0. Chng minh rng (abc)a+b+c

3 aabbcc.

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Chuyn 5. Hm S Ly Tha. Hm S M & Hm S Lgarit

3. Hm S Ly Tha. Hm S M & Hm S LgaritA. Kin Thc Cn Nh

1. Hm s lu tha. Dng: y = x ( R). Tp xc nh:

Nu nguyn dng th D = R.Nu = 0 hoc nguyn m th D = R\ {0}.Nu khng nguyn th D = (0; +).

o hm: y = x1. Tnh cht: (Xt trn (0; +))

> 0: Hm s lun ng bin. < 0: Hm s lun nghch bin.

O O

y y

x x

> 0 < 0

2. Hm s m. Dng: y = ax (0 < a 6= 1). Tp xc nh: D = R. o hm: y = ax ln a. Tnh cht:

a > 1: Hm s lun ng bin.a < 1: Hm s lun nghch bin.

O O

y y

x x

a > 1 0 < a < 1

1 1

3. Hm s lgarit. Dng: y = loga x (0 < a 6= 1). Tp xc nh: D = (0; +). o hm: y = 1x ln a . Tnh cht:

a > 1: Hm s lun ng bin.a < 1: Hm s lun nghch bin.

O O

y y

x x

a > 1 0 < a < 1

1 1

4. o hm ca hm s ly tha, m v lgarit. (x) = x1. (u) = u1.u. (ex) = ex. (eu) = ueu. (ax) = ax ln a. (au) = uau ln a. (lnx) = 1

x. (lnu) = u

u. (logax) =

1

x ln a. (logau) =

u

u ln a.

B. Bi Tp

5.15. Tm tp xc nh ca cc hm s saua) y =

(x2 2)2. b) y = (2 x2) 27 . c) y = (x2 x 2)2.

d) y = log2 (5 2x). e) y = log3(x2 2x). f) y = log0,4 3x+21x .

5.16. Tnh o hm ca cc hm s saua) y =

(3x2 4x+ 1)2. b) y = 3x2 lnx+ 4 sinx. c) y = 2xex + 3 sin 2x.

d) y = log(x2 + x+ 1

). e) y = ln e

x

1+ex . f) y =(x2 14

)e2x.

g) y =(e4x + 1 lnx)pi. h) y = 2 ln x+14 ln x5 . i) y = ln (2ex + ln (x2 + 3x+ 5)).

5.17. Tm gi tr ln nht v gi tr nh nht ca cc hm s saua) y = x e2x trn [0; 1]. b) y = e2x 2ex trn [1; 2]. c) y = (x+ 1) ex trn [1; 2].d) y = ln

(3 + 2x x2) trn [0; 2]. e) y = ln (4 3x2 x4). f) y = x2 ln (1 2x) trn [2; 0].

g) y = x2ex trn [0; ln 8]. h) y = x2 lnx trn [1; e]. i) y = 5x + 51x trn [0; log58].

4. Phng Trnh & Bt Phng Trnh MA. Kin Thc Cn Nh

1. Phng trnh m c bn. Dng: ax = b (0 < a 6= 1). Cch gii:

b 0: Phng trnh v nghim.b > 0: ax = b x = logab.

2. Bt phng trnh m c bn. Dng: ax > b (0 < a 6= 1). Cch gii:

b 0: S = R.b > 0, a > 1: ax > b x > logab.

0 < a < 1: ax > b x < logab.Lu . Cc dng ax b; ax < b; ax b da vo du c cch gii tng ng.

31

Nguyn Minh Hiu

B. Phng Phng Gii C Bn

a v cng c s. t n ph. Ly lgarit hai v. S dng tnh n iu ca hm s m.

C. Bi Tp

5.18. Gii cc phng trnh saua) 22x1 = 3. b) 2x

2x = 4.c) 2x

2x+8 = 413x. d) 3x.2x+1 = 72.e) 32x1 + 32x = 108. f) 2x + 2x+1 + 2x+2 = 3x + 3x1 + 3x2.

g)(3 + 2

2)x+1

=(3 22)2x+8. h) (5 26)x23x+2 (5 + 26) 1x22 = 0.

5.19. Gii cc bt phng trnh saua) 2x

2+3x < 4. b) 3x+2 + 3x1 28.c) 2x+2 2x+3 2x+4 > 5x+1 5x+2. d) 2x + 2x+1 + 2x+2 < 3x + 3x1 + 3x2.e) x2x1 < xx

2

. f)(

5 + 2)x1 (5 2) x1x+1 .

g) 32x+5x1 > 0, 25.128

x+17x3 . h) 2x

2

.7x2+1 < 7.142x

24x+3.

5.20. Gii cc phng trnh saua) 64x 8x 56 = 0. b) (TN-08) 32x+1 9.3x + 6 = 0.c) 22+x 22x = 15. d) (TN-07) 7x + 2.71x 9 = 0.e) (D-03) 2x

2x 22+xx2 = 3. f) 32x+1 = 3x+2 +

1 6.3x + 32(x+1).5.21. Gii cc bt phng trnh sau

a) 4x 3.2x + 2 > 0. b) 32.4x + 1 < 18.2x.c) 5x + 51x > 6. d)

(2 +

3)x

+(23)x > 4.

5.22. Gii cc phng trnh saua)(5 26)x + (5 + 26)x = 10. b) (B-07) (2 1)x + (2 + 1)x 22 = 0.

c)(7 + 3

5)x

+ 5.(7 35)x = 6.2x. d) (5 + 26)x + (5 26)x = 10.

e)(7 + 4

3)x 3(23)x + 2 = 0. f) (26 + 153)x + 2(7 + 43)x 2(23)x = 1.

5.23. Gii cc phng trnh saua) 3.4x 2.6x = 9x. b) 2.16x+1 + 3.81x+1 = 5.36x+1.c) 4x+

x22 5.2x1+

x22 6 = 0. d) 5.2x = 7

10x 2.5x.

e) 27x + 12x = 2.8x. f) (A-06) 3.8x + 4.12x 18x 2.27x = 0.5.24. Gii cc bt phng trnh sau

a) 27x + 12x < 2.8x. b) 252xx2+1 + 92xx

2+1 34.152xx2 .c) 9

1x 13.6 1x1 + 4 1x < 0. d) 9x 3x+1 + 2 > 3x 9.

e) 45x

52x5x+1+6 1. f) 47.5x

52x+112.5x+4 23 .

5.25. Gii cc phng trnh saua) 12 + 6x = 4.3x + 3.2x. b) 52x+1 + 7x+1 175x 35 = 0.c) 2x

25x+6 + 21x2

= 2.265x + 1. d) (D-06) 2x2+x 4.2x2x 22x + 4 = 0.

e) 4x2+x + 21x

2

= 2(x+1)2

+ 1. f) x2.2x1 + 2|x3|+6 = x2.2|x3|+4 + 2x+1.

5.26. Gii cc bt phng trnh saua) 12 + 6x > 4.3x + 3.2x. b) 4x

2+x + 21x2 2(x+1)2 + 1.

c) 52x+1 + 6x+1 > 30 + 5x.30x. d) 52x103x2 4.5x5 < 51+3

x2.

5.27. Gii cc phng trnh saua) 3x = 11 x. b) 2x = x+ 1.c) 3x + 4x = 5x. d) 1 + 8

x2 = 3x.

e) 5x22x+2 + 4x

22x+3 + 3x22x+4 = 48. f) 2

3 x = x2 + 8x 14.

5.28. Gii cc phng trnh saua) 4x + (2x 17) .2x + x2 17x+ 66 = 0. b) 9x + 2 (x 2) .3x + 2x 5 = 0.c) 9x

2

+(x2 3) .3x2 2x2 + 2 = 0. d) 32x (2x + 9) .3x + 9.2x = 0.

32

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Chuyn 5. Hm S Ly Tha. Hm S M & Hm S Lgarit

5.29. Gii cc phng trnh saua) 22x 2x + 6 = 6. b) 32x +3x + 7 = 7.c) 27x + 2 = 3 3

3x+1 2. d) 7x1 = 6log7 (6x 5) + 1.

5.30. Gii cc phng trnh saua) 2x

2

= 3x. b) 2x24 = 3x2.

c) 5x.8x1x = 500. d) 8

xx+2 = 4.34x.

5.31. Gii cc phng trnh saua) 3x

2

= cos 2x. b) 2|x| = sinx.c) 2x1 + 2x

2 x.2x1 2x2x = (x 1)2. d) 22x+1 + 232x = 8log3(4x24x+4) .

5. Phng Trnh & Bt Phng Trnh LgaritA. Kin Thc Cn Nh

1. Phng trnh lgarit c bn. Dng: logax = b (0 < a 6= 1). Cch gii: logax = b x = ab.

2. Bt phng trnh lgarit c bn. Dng: logax > b (0 < a 6= 1). Cch gii: a > 1: logax > b x > ab.

0 < a < 1: logax > b 0 < x < ab.Lu . Cc dng logax b; logax < b; logax b da vo du c cch gii tng ng.

B. Phng Phng Gii C Bn

a v cng c s. t n ph. S dng tnh n iu ca hm s lgarit.

C. Bi Tp

5.32. Gii cc phng trnh saua) log3 (x 2) = 2. b) log3 (5x+ 3) = log3 (7x+ 5).c) log2

(x2 1) = log 1

2(x 1). d) log2x+ log2 (x 2) = 3.

e) log2(x2 + 8

)= log2x+ log26. f) log3 (x+ 2) + log3 (x 2) = log35.

g) log3x+ log4x = log5x. h) log2x+ log3x+ log4x = log20x.

5.33. Gii cc bt phng trnh saua) log8 (4 2x) 2. b) log3

(x2 + 2

)+ log 1

3(x+ 2) < 0.

c) log 15

(3x 5) > log 15

(x+ 1). d) log2 (x+ 3) < log4 (2x+ 9).

5.34. Gii cc phng trnh saua) log2

(x2 + 3x+ 2

)+ log2

(x2 + 7x+ 12

)= log224. b) log

(x3 + 8

)= log (x+ 58) + 12 log

(x2 + 4x+ 4

).

c) 12 log

2 (x+ 3) +14 log4(x 1)8 = log24x. d) 32 log 14 (x+ 2)

2 3 = log 14(4 x)3 + log 1

4(x+ 6)

3.

e) log2x+ 1 log 1

2(3 x) log8(x 1)3 = 0. f) log 12 (x 1) + log 12 (x+ 1) log 12 (7 x) = 1.

g) log2(8 x2)+ log 1

2

(1 + x+

1 x) 2 = 0. h) log2 (4x + 15.2x + 27) + 2log2 14.2x3 = 0.

5.35. Gii cc phng trnh saua) log2

(xx2 1)+ 3log2 (x+x2 1) = 2. b) (A-08) log2x1 (2x2 + x 1)+logx+1(2x 1)2 = 4.

c) log2(xx2 1) .log3 (x+x2 1) = log6 (xx2 1).

5.36. Gii cc bt phng trnh saua) (A-07) 2log3 (4x 3) + log 13 (2x+ 3) 2. b) log 12x+ 2log 14 (x 1) + log26 0.c) (D-08) log 1

2

x23x+2x 0. d) log0,5 x+12x1 > 1.

e)log2

(3.2x1 1)x

1. f)log2 (1 3log27x) 1

log2x< 0.

g) (B-02) logx [log3 (9x 72)] 1. h) x 1

log3 (9 3x) 3 1.

5.37. Gii cc bt phng trnh saua) (B-08) log0,7

(log6

x2+xx+4

)< 0. b) log 1

2log3

x+1x1 0.

c) log3log43x1x+1 log 13 log 14

x+13x1 . d) log 13 log5

(x2 + 1 + x

)> log3log 15

(x2 + 1 x).

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Nguyn Minh Hiu

5.38. Gii cc phng trnh saua) log22 x 3log2x+ 2 = 0. b) log 12x+ log

22 x = 2.

c) 2log2x log3x = 2 log x. d) log2x3 20 logx+ 1 = 0.e)

log3x+

4 log3x = 2. f) log2 (2x + 1) .log2(2x+1 + 2

)= 2.

g) log3 (3x + 1) .log3(3x+2 + 9

)= 3. h) log2 (5x 1) .log4 (2.5x 2) = 1.

5.39. Gii cc bt phng trnh saua) log22 (2x+ 1) 3 log (2x+ 1) + 2 > 0. b) log29 (x 1) 3log3 (x 1) + 1 0.c) logx14 1 + log2 (x 1). d) log2 (2x 1) log 12

(2x+1 2) > 2.

e) log4 (19 2x) log2 192x

8 1. f) log5 (4x + 144) 4log52 < 1 + log5(2x2 + 1

).

5.40. Gii cc bt phng trnh saua)

log2x+

logx2 43 . b) 3

log 12x+ log4x

2 2 > 0.c)

log22 x+ log 12x2 3 > 5 (log4x2 2). d) log22 x+ log2x4 8 > log2 x24 .

5.41. Gii cc bt phng trnh saua) log2x64 + logx216 3. b) logx (125x) .log25x > 32 + log25 x.c) (C-2012) log2(2x). log3(3x) > 1. d) log 13x+

1 4 log21

2x < 1.

5.42. Gii cc phng trnh saua) x+ 2.3log2x = 3. b) x2 + 3log2x = xlog25.c) xlog29 = x2.3log2x xlog23. d) log2

(x+ 3log6x

)= log6x.

5.43. Gii cc phng trnh saua) log22 x+ (x 4) log2x x+ 3 = 0. b) log22 (x+ 1) + (x 5) log2 (x+ 1) 2x+ 6 = 0.c) log2

(x2 + 1

)+(x2 5) log (x2 + 1) 5x2 = 0. d) (x+ 2) log23 (x+ 1)+4 (x+ 1) log3 (x+ 1)16 = 0.

5.44. Gii cc phng trnh saua) log2 (1 +

x) = log3x. b) log7x = log3 (2 +

x).

c) 3log3 (1 +x+ 3x) = 2log2

x. d) log 1

2(3 + |x|) = 2|x| 4.

e) log2(x2 4)+ x = log2 [8 (x+ 2)]. f) 4 (x 2) [log2 (x 3) + log3 (x 2)] = 15 (x+ 1).

5.45. Gii cc bt phng trnh saua) 3x > 11 x. b) 1 +15x 4x.c) 1 + 2x+1 + 3x+1 < 6x. d) 4log x+1 6log x > 2.3log x2+2.e) log7x < log3 (

x+ 2). f) log2 (2x + 1) + log3 (4x + 2) 2.

6. H Phng Trnh M & Lgarit5.46. Gii cc h phng trnh sau

a){

3y+1 2x = 54x 6.3y + 2 = 0 . b) (D-02)

{23x = 5y2 4y4x+2x+1

2x+2 = y.

c) (A-09){

log2(x2 + y2

)= 1 + log2 (xy)

3x2xy+y2 = 81

. d) (B-2010){

log2 (3y 1) = x4x + 2x = 3y2

.

5.47. Gii cc h phng trnh sau

a){

log3 (x+ 2) < 3log 1

2

(x2 + 2x 8) log 1

216

. b) (A-04){

log 14

(y x) log4 1y = 1x2 + y2 = 25

.

c) (D-2010){x2 4x+ y + 2 = 02log2 (x 2) log2y = 0

. d) (B-05){

x 1 +2 y = 13log99x

2 log3y3 = 3 .

5.48. Gii cc h phng trnh sau

a){

3x 3y = y xx2 + xy + y2 = 12

. b){x3 y3 = 2y 2x(x4 + 1

) (y2 + y 1)+ x (y 2) = 1 .

c){x+x2 2x+ 2 = 3y1 + 1

y +y2 2y + 2 = 3x1 + 1 . d)

{ln (1 + x) ln (1 + y) = x yx2 12xy + 20y2 = 0 .

5.49. (D-06) Chng minh vi mi a > 0, h phng trnh{ex ey = ln (1 + x) ln (1 + y)y x = a c nghim duy nht.

34

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Chuyn 6

Phng Php Ta Trong Khng Gian

1. Ta Trong Khng GianA. Kin Thc Cn Nh

1. Ta trong khng gian.

Hai vect bng nhau: a = b a1 = b1a2 = b2

a3 = b3

.

Cc php ton vect: a b = (a1 b1; a2 b2; a3 b3); ka = (ka1; ka2; ka3). Tch v hng ca hai vect: a .b = a1b1 + a2b2 + a3b3. Hai vect vung gc: a b a .b = 0. di vect: |a | =

a21 + a

22 + a

23.

Gc gia hai vect: cos(a ;b ) = a .b

|a | .b .

Ta vect: AB = (xB xA; yB yA; zB zA). Khong cch gia hai im: AB =

AB = (xB xA)2 + (yB yA)2 + (zB zA)2. Tnh cht trung im: I l trung im ca AB I

(xA + xB

2;yA + yB

2;zA + zB

2

).

Tnh cht trng tm: G l trng tm ABC G(xA + xB + xC

3;yA + yB + yC

3;zA + zB + zC

3

).

2. Tch c hng ca hai vct.

nh ngha.[a ,b ] = (a2 a3b2 b3

; a3 a1b3 b1 ; a1 a2b1 b2

). Tnh cht.[a ,b ]a ; [a ,b ]b . [a ,b ] = |a | . b . sin(a ,b ).[a ,b ] = 0 a ,b cng phng. [a ,b ].c = 0 a ,b ,c ng phng.

ng dng. Din tch tam gic: SABC = 12

[AB,AC]. Th tch t din: VABCD = 16 [AB,AC] .AD. Th tch hnh hp: VABCD.ABCD =

[AB,AD] .AA.3. Phng trnh mt cu. Dng 1: (x a)2 + (y b)2 + (z c)2 = R2 (R > 0).C tm I (a; b; c) v bn knh R =

R2.

Dng 2: x2 + y2 + z2 2ax 2by 2cz + d = 0 (a2 + b2 + c2 > d).C tm I (a; b; c) v bn knh R =

a2 + b2 + c2 d.

Lu . im M thuc mt cu R = IM .

B. Bi Tp

6.1. Trong khng gian Oxyz, cho ba vect a (5; 7; 2) ,b (3; 0; 4) v c (6; 1;1).a) Hy tm cc vect sau: m = 3a 2b +c ;n = 5a + 6b + 4c ;p = 12a 13

b + 16

c .

35

Nguyn Minh Hiu

b) Tnh: |a | ;b ; a b ;a .b ; [a ,b ].

c) Tm x sao cho a + 3b 2x = 0 .d) Tm u, v vect y (1;u; v) cng phng vi vect a + 2b .

6.2. Trong khng gian Oxyz, cho ba vect a (1; 0;2) ,b (1; 2;1) v c (0; 3;2).a) Tm vect u bit 2a +b 3c 2u = 0 . b) Tnh

a +b +c .c) Tm a

(b 2c

);[a ,b ]. d) Tm vect u bit ua ;ub v |u | = 21.

6.3. Trong khng gian Oxyz, cho ba im A (1; 0;2) , B (2; 1;1) , C (1;2; 2).a) Chng minh A,B,C khng thng hng. b) Tnh chu vi tam gic ABC.c) Tm ta D ABCD l hnh bnh hnh. d) Tm ta trng tm tam gic ABC.

6.4. Trong khng gian Oxyz, cho ba im A (1;2; 3) , B (0; 3; 1) , C (4; 2; 2).a) Tnh

AB.AC. b) Tnh cos BAC. c) Tnh

[AB,

AC].

6.5. Trong khng gian Oxyz, cho ba im A (1; 0; 3) , B (2; 2; 4) , C (0; 3;2).a) Chng minh tam gic ABC vung ti A.b) Tm tm v bn knh ng trn ngoi tip tam gic ABC.c) Tnh din tch tam gic ABC.

6.6. Trong khng gian Oxyz, cho tam gic ABC c A (1; 1; 3) , B (1; 3; 2) , C (1; 2; 3). Tnh din tch tam gicABC v th tch t din OABC.

6.7. Trong khng gian Oxyz, cho hai im A (3;2; 6) , B (2; 4; 4). Hy tnh di ng cao OH ca tam gicOAB.

6.8. Trong khng gian Oxyz, cho tam gic ABC c A (0; 4; 1) , B (1; 0; 1) , C (3; 1;2) .. Tm to trc tm tamgic ABC.

6.9. Trong khng gian Oxyz, cho ba im A (1; 1; 1) , B (1; 1; 0) , C (3; 1;1). Tm imM thuc mt phng (Oxz)sao cho M cch u A,B,C.

6.10. Trong khng gian Oxyz, cho hai im A (1; 6; 6) , B (3;6;2). Tm im M thuc mt phng (Oxy) saocho AM +BM l ngn nht.

6.11. Trong khng gian Oxyz, cho ba im A (2; 5; 3) , B (3; 7; 4) , C (x, y, 6). Tm x, y A,B,C thng hng.

6.12. Trong khng gian Oxyz, cho bn im A (1; 1; 1) , B (2; 3; 4) , C (6; 5; 2) , D (7, 7, 5). Chng minh rng A,B,C,Dl bn nh ca hnh bnh hnh. Tnh din tch hnh bnh hnh .

6.13. Trong khng gian Oxyz, cho t din ABCD c A (2; 1;1) , B (3; 0; 1) , C (2;1; 3) v D thuc trc Oy. Tmta nh D, bit th tch t din ABCD bng 5.

6.14. Tm tm v bn knh ca cc mt cu saua) (x 3)2 + (y + 2)2 + (z + 1)2 = 9. b) x2 + y2 + z2 + 2x+ 4y 6z + 9 = 0.c) x2 + y2 + z2 + y 5z + 1 = 0. d) 3x2 + 3y2 + 3z2 6x+ 8y + 15z 3 = 0.

6.15. Lp phng trnh mt cu trong cc trng hp saua) C tm I (1; 2;3) v qua M (2; 0;1).b) C ng knh AB bit A (3; 2;1) v B (1; 1; 2).c) Ngoi tip t din OABC bit A (2; 0; 0) , B (0;1; 0) v C (0; 0; 3).d) Ngoi tip t din ABCD bit A (1; 2; 1) , B (3;1; 2) , C (2; 1; 2) v D (1; 1; 3).e) C tm nm trn mt phng (Oyz) v qua ba im A (0; 8; 0) , B (4; 6; 2) , C (0; 12; 4).

2. Phng Trnh Mt PhngA. Kin Thc Cn Nh

1. Vect php tuyn ca mt phng.nh ngha: Vect n 6= 0 c gi vung gc vi () gi l vect php tuyn ca mt phng ().Lu . Mt mt phng c v s vect php tuyn cng phng. Nu hai vect a ,b khng cng phng v c gi song song hoc cha trong () th n =

[a ,b ].36

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Chuyn 6. Phng Php Ta Trong Khng Gian

2. Phng trnh tng qut ca mt phng.Dng: () : Ax+By + Cz +D = 0 (A,B,C khng ng thi bng 0).

Nhn xt. Mt phng Ax+By + Cz +D = 0 c vect php tuyn n (A;B;C). Ly x0; y0 tu z0 ta c im M (x0; y0; z0) ().Mt phng quaM (x0; y0; z0) v c vect php tuynn (A;B;C) c PT: A (x x0)+B (y y0)+C (z z0) = 0. Mt phng qua A (a; 0; 0) , B (0; b; 0) v C (0; 0; c) c phng trnh x

a+y

b+z

c= 1 gi l PT on chn.

Mt phng (Oxy) , (Oxz) , (Oyz) ln lt c phng trnh z = 0, y = 0, x = 0. Mt phng () tip xc vi mt cu (S) d (I; ()) = R.

3. V tr tng i ca hai mt phng.Cho hai mt phng (1) : A1x+B1y + C1z +D1 = 0 v (2) : A2x+B2y + C2z +D2 = 0. Ta c:

(1) // (2){ n1 = kn2D1 6= kD2 . (1) (2)

{ n1 = kn2D1 = kD2

.

(1) ct (2) n1 6= kn2. (1) (2) n1.n2 = 0.4 . Khong cch.

T mt im n mt mt phng: d (M ; ()) = |AxM +ByM + CzM +D|A2 +B2 + C2

.

Gia hai mt phng song song: d (() ; ()) = d (M ; ()) (M ()).

B. Bi Tp

6.16. Lp phng trnh mt phng (P ) trong cc trng hp saua) i qua ba im A (1; 0; 0) , B (0;2; 0) , C (0; 0; 3).b) i qua ba im A (2;1; 3) , B (4; 2; 1) , C (1; 2; 3).c) i qua im M (2;1; 2) v song song vi mt phng () : 2x y + 3z + 4 = 0.d) i qua M (1; 2; 3) v vung gc AB. Bit A (1; 0; 2) , B (3; 2; 1).e) i qua hai im A (3; 1;1) , B (2;1; 4) v vung gc vi mt phng () : 2x y + 3z + 1 = 0.f) i qua M (2; 3;1) v vung gc vi hai mt phng () : x+ 2y + 2z + 1 = 0; () : 2x+ 3y + z = 0.g) i qua hai im M (1; 2; 3) , N (2;2; 4) v song song vi trc Oy.h) Trung trc ca Ab, bit A (4;1; 5) , B (2; 3; 1).i) Song song vi () : 4x+ 3y 12z + 1 = 0 v tip xc vi (S) : x2 + y2 + z2 2x 4y 6z 2 = 0.

6.17. Xt v tr tng i ca cc cp mt phng saua) () : x 2y + 3z 3 = 0; () : 2x y + z 1 = 0.b) () : 2x y + 2z + 1 = 0; () : 4x+ 2y 4z 1 = 0.c) () : 3x y + 2z + 1 = 0; () : 6x 2y + 4z + 2 = 0.

6.18. Tnh cc khong cch saua) Gia M (2;3; 1) v () : 2x+ 2y + z + 3 = 0.b) Gia A (4; 1; 5) v () : x+ 7y 2z + 1 = 0.c) Gia () : 2x y + 2z + 1 = 0 v () : 4x 2y + 4z 3 = 0.

6.19. (TN-06) Trong khng gian Oxyz, cho ba im A (2; 0; 0) , B (0; 3; 0) , C (0; 0; 6).a) Vit phng trnh mt phng qua ba im A,B,C.b) Gi G l trng tm tam gic ABC. Vit phng trnh mt cu ng knh OG.

6.20. (TN-07) Trong khng gian Oxyz, cho im E (1;4; 5) , F (3; 2; 7).a) Vit phng trnh mt cu qua F v c tm E.b) Vit phng trnh mt phng trung trc ca EF .

6.21. (C-09) Trong khng gian Oxyz, cho (P1) : x + 2y + 3z + 4 = 0 v (P2) : 3x + 2y z + 1 = 0. Vit phngtrnh mt phng (P ) i qua A (1; 1; 1) v vung gc vi hai mt phng (P1) , (P2).

6.22. Trong khng gian Oxyz, cho t din ABCD c A (5; 1; 3) , B (1; 6; 2) , C (5; 0; 4) , D (4; 0; 6).a) Vit phng trnh cc mt phng (ACD) v (BCD).b) Vit phng trnh mt phng () cha cnh AB v song song vi cnh CD.

6.23. Trong khng gian Oxyz, cho bn im A (2; 6; 3) , B (1; 0; 6) , C (0; 2;1) , D (1; 4; 0).a) Vit phng trnh mt phng (BCD). Suy ra ABCD l mt t din.b) Tnh chiu cao AH ca t din ABCD.c) Vit phng trnh mt phng () cha AB v song song vi CD.

37

Nguyn Minh Hiu

6.24. (C-2011) Trong khng gianOxyz, cho hai im A (1; 2; 3) , B (1; 0;5) v mt phng (P ) : 2x+y3z4 = 0.Tm im M thuc (P ) sao cho ba im A,B,M thng hng.

6.25. Trong khng gian Oxyz, vit phng trnh mt phng () song song vi mt phng () : 4x+3y12z+1 = 0v tip xc vi mt cu (S) : x2 + y2 + z2 2x 4y 6z 2 = 0.6.26. (D-04) Trong khng gian Oxyz, cho ba im A (2; 0; 1) , B (1; 0; 0) , C (1; 1; 1) v (P ) : x+ y + z 2 = 0. Vitphng trnh mt cu i qua ba im A,B,C v c tm thuc (P ).

6.27. (B-2012) Trong khng gian Oxyz, cho A(0; 0; 3),M(1; 2; 0). Vit phng trnh mt phng (P ) qua A v ctcc trc Ox,Oy ln lt ti B,C sao cho tam gic ABC c trng tm thuc ng thng AM .

6.28. (A-2011) Trong khng gian Oxyz, cho mt cu (S) : x2 + y2 + z2 4x 4y 4z = 0 v im A (4; 4; 0). Vitphng trnh mt phng (OAB), bit im B thuc (S) v tam gic OAB u.

6.29. (A-2011) Trong khng gian Oxyz, cho hai im A (2; 0; 1) , B (0;2; 3) v mt phng (P ) : 2x y z+ 4 = 0.Tm ta im M thuc (P ) sao cho MA = MB = 3.

6.30. (B-08) Trong khng gian Oxyz, cho ba im A (0; 1; 2) , B (2;2; 1) , C (2; 0; 1).a) Vit phng trnh mt phng qua ba im A,B,C.b) Tm to im M thuc (P ) : 2x+ 2y + z 3 = 0 sao cho MA = MB = MC.

6.31. (D-2010) Trong khng gian Oxyz, cho (P ) : x+ y + z 3 = 0 v (Q) : x y + z 1 = 0. Vit phng trnhmt phng (R) vung gc vi (P ) v (Q) sao cho khong cch t O n (R) bng 2.

6.32. (B-09) Trong khng gian Oxyz, cho t din ABCD c cc nh A (1; 2; 1) , B (2; 1; 3) , C (2;1; 1) , D (0; 3; 1).Vit phng trnh mt phng (P ) i qua A,B sao cho khong cch t C n (P ) bng khong cch t D n (P ).

6.33. (B-07) Trong khng gian Oxyz, cho mt cu (S) : x2 + y2 + z2 2x + 4y + 2z 3 = 0 v mt phng(P ) : 2x y + 2z 14 = 0.

a) Vit phng trnh (Q) cha trc Ox v ct (S) theo ng trn c bn knh bng 3.b) Tm im M thuc mt cu (S) sao cho khong cch t M n (P ) l ln nht.

6.34. Trong khng gian Oxyz, cho ba im A (0; 1; 1) , B (2;1; 1) , C (4; 1; 1) v mt phng (P ) : x+ y+ z 6 = 0.Tm im M trn (P ) sao cho

MA+ 2MB +MC t gi tr nh nht.6.35. (A-03) Trong khng gian Oxyz, cho hnh hp ch nht ABCD.ABC D c A trng gc to O, B (a; 0; 0),D (0; a; 0) , A (0; 0; b) , (a > 0, b > 0). Gi M l trung im cnh CC .

a) Tnh th tch khi t din BDAM .b) Xc nh t s ab (A

BD) vung gc vi (MBD).

3. Phng Trnh ng ThngA. Kin Thc Cn Nh

1. Vect ch phng ca ng thng.nh ngha: Vect u 6= 0 c gi song song hoc trng vi gi l vect ch phng ca .Lu . ng thng c v s vect ch phng cng phng vi nhau.

2. Phung trnh tham s ca ng thng.

ng thng qua M (x0; y0; z0) v nhn u (a1; a2; a3) lm vect ch phng c PTTS: x = x0 + a1ty = y0 + a2t

z = z0 + a3t().

Nhn xt. ng thng qua M (x0; y0; z0) v c vect ch phng u (a1; a2; a3). Nu a1a2a3 6= 0 th cn vit di dng x x0

a1=y y0a2

=z z0a3

gi l dng chnh tc.

Nu song song vi () v M th d (; ()) = d (M ; ()).3. V tr tng i gia hai ng thng.

d d [u ,u] = [u ,M0M 0] = 0 . d v d ct nhau

[u ,u] 6= 0[u ,u] .M0M 0 = 0 .

38

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Chuyn 6. Phng Php Ta Trong Khng Gian

d//d [u ,u] = 0[u ,M0M 0] 6= 0 . d v d cho nhau

[u ,u] .M0M 0 6= 0.4. V tr tng i gia ng thng v mt phng.

Cho ng thng :

x = x0 + atx = y0 + btz = z0 + ct

v mt phng () :Ax+By + Cz +D = 0.

S giao im ca v () l s nghim phng trnh A (x0 + at) +B (y0 + bt) + C (z0 + ct) +D = 0 (1). //() (1) v nghim. () (1) c v s nghim. ct () (1) c mt nghim. () u = kn .

B. Bi Tp

6.36. Lp phng trnh ng thng d trong cc trng hp saua) i qua A (2; 1;1) v c vect ch phng u = (2; 3; 2).b) i qua hai im A (1; 2; 3) , B (5; 4; 4).c) i qua A (3; 1; 2) v vung gc vi () : x 2y + 3z + 1 = 0.d) i qua M (2; 1;3) v song song vi ng thng : x 1

2=y + 3

3=z

4.

e) i qua M (3; 1; 4) v song song vi giao tuyn ca () : 3x+ 2y 5z + 1 = 0; () : x 4y + 3z + 2 = 0.f) Giao tuyn ca () : x+ z 1 = 0; () : 2x 2y + 3z + 1 = 0.

6.37. (C-09) Trong khng gian Oxyz, cho tam gic ABC c A (1; 1; 0) , B (0; 2; 1) v trng tm G (0; 2;1). Vitphng trnh ng thng i qua C v vung gc vi (ABC).

6.38. (TN-08) Trong khng gian Oxyz, cho im A (2;1; 3) v (P ) : x 2y 2z 10 = 0. Tnh khong cch t An (P ). Vit phng trnh ng thng qua A v vung gc vi (P ).

6.39. (D-2011) Trong khng gian Oxyz, cho :x 1

2=y 3

4=z

1v (P ) : 2x y + 2z = 0. Vit phng trnh

mt cu c tm thuc ng thng , bn knh bng 1 v tip xc vi mt phng (P ).

6.40. Trong khng gian Oxyz, cho cc im A (4;6; 3) , B (5;7; 3). Gi d l ng thng qua A v vung gc vimt phng (P ) : 8x+ 11y + 2z 3 = 0. Tm im C thuc d sao cho ABC vung ti B.

6.41. (A-2012) Trong khng gian Oxyz, cho ng thng d :x+ 1

2=y

1=z 2

1, mt phng (P ) : x+y2z+5 = 0

v im A(1;1; 2). Vit phng trnh ng thng ct d v (P ) ln lt ti M v N sao cho A l trung imca on thng MN .

6.42. (D-2012) Trong khng gianOxyz, cho ng thng d :x 1

2=y + 1

1 =z

1v hai imA (1;1; 2) , B (2;1; 0).

Xc nh ta im M thuc d sao cho tam gic AMB vung ti M .

6.43. (A-2012) Trong khng gian Oxyz, cho ng thng d :x+ 1

1=y

2=z 2

1v im I(0; 0; 3). Vit phng

trnh mt cu (S) c tm I v ct d ti hai im A,B sao cho tam gic IAB vung ti I.

6.44. Xt v tr tng i ca ng thng v mt phng sau

a) d :x 12

4=y 9

3=z 1

1v () : 3x+ 5y z 2 = 0.

b) d :x 1

1=y 1

2=z 23 v () : x+ y + z 4 = 0.

6.45. Xt v tr tng i ca cc cp ng thng sau

a) d :x 1

1=y

2=z 31 v d

:x 2

2=y 3

4=z 52 .

b) d :x 31 =

y 41

=z 52 v d

:x 23 =

y 53

=z 36 .

c) d :x 1

1=y 2

3=z 31 v d

:x 22 =

y + 2

1=z 1

3.

d) d :x 1

2=y + 1

3=z 5

1v d :

x 13

=y + 2

2=z + 1

2.

6.46. (TN-09) Trong khng gian Oxyz, cho mt cu (S) : (x 1)2 + (y 2)2 + (z 2)2 = 36 v mt phng(P ) : x+ 2y + 2z + 18 = 0.

a) Xc nh to tm T v bn knh ca (S). Tnh khong cch t T n (P ).b) Vit phng trnh tham s ca ng thng d qua T v vung gc (P ). Tm to giao im ca d v (P ).

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Nguyn Minh Hiu

6.47. Trong khng gian Oxyz, cho mt phng () : 3x+ 5y z2 = 0 v ng thng d : x 124

=y 9

3=z 1

1.

a) Tm giao im M ca d v ().b) Vit phng trnh mt phng () cha M v vung gc vi d.

6.48. Trong khng gian Oxyz, cho hai ng thng d :x 11 =

y

1=

z

1 v d :x

2=y + 1

1=z

1.

a) Chng minh d v d cho nhau.b) Vit phng trnh mt phng () cha d v song song vi d.

6.49. (C-2012) Trong khng gian Oxyz, cho hai ng thng d1 :

x = ty = 2tz = 1 t

, d2 :

x = 1 + 2sy = 2 + 2sz = s

. Chng minh

d1 v d2 ct nhau. Vit phng trnh mt phng cha hai ng thng d1, d2.

6.50. (B-06) Trong khng gian Oxyz cho im A (0; 1; 2) v d1 :x

2=y 1

1=z + 1

1 , d2 :x 1

1=y + 1

2 =z 2

1.

a) Vit phng trnh mt phng (P ) qua A v song song vi d1, d2.b) Tm M thuc d1, N thuc d2 sao cho A,M,N thng hng.

6.51. (D-03) Trong khng gian Oxyz, cho d l giao tuyn ca hai mt phng (P ) : x + 3ky z + 2 = 0 v(Q) : kx y + z + 1 = 0. Tm k d vung gc vi () : x y 2z + 5 = 0.6.52. (D-02) Trong khng gian Oxyz, cho (P ) : 2x y + 2 = 0 v d l giao tuyn hai mt phng () : (2m+ 1)x+(1m) y +m 1 = 0 v () : mx+ (2m+ 1) z + 4m+ 2 = 0. Xc nh m d song song vi (P ).6.53. (D-09) Trong khng gian Oxyz, cho A (2; 1; 0) , B (1; 2; 2) , C (1; 1; 0) v mt phng (P ) : x + y + z 20 = 0.Xc nh to im D thuc ng thng AB sao cho ng thng CD song song vi mt phng (P ).

6.54. Lp phng trnh ng vung gc chung ca d1 :x 7

1=y 3

2=z 91 v d2 :

x 37 =

y 12

=z 1

3.

6.55. Vit phng trnh ng thng qua A (1;1; 1) v ct d : x 12

=y

1=z 31 , d

: x1 =y+12 =

z21 .

6.56. Vit phng trnh ng thng vung gc vi (Oxz) v ct d :

x = ty = 4 + tz = 3 t

, d :

x = 1 2t

y = 3 + tz = 4 5t

.

6.57. (B-04) Trong khng gian Oxyz, cho A (4;2; 4) v d : x+ 32

=y 11 =

z + 1

4. Vit phng trnh ng

thng qua A, ct v vung gc vi d.

6.58. (D-09) Trong khng gian Oxyz, cho :x+ 2

1=y 2

1=

z

1 v (P ) : x+ 2y 3z+ 4 = 0. Vit phng trnhng thng d nm trong (P ) sao cho d ct v vung gc vi .

6.59. (A-07) Trong khng gian Oxyz, cho hai ng thng d1 :x

2=y 11 =

z + 2

1v d2 :

x = 2 + 2ty = 1 + tz = 3

.

a) Chng minh d1 v d2 cho nhau.b) Vit phng trnh d vung gc vi (P ) : 7x+ y 4z = 0 v ct hai ng thng d1, d2.

6.60. (C-2012) Trong khng gian Oxyz, cho ng thng d :x 21 =

y + 1

1 =z + 1

1v mt phng (P ) :

2x + y 2z = 0. ng thng nm trong (P ) vung gc vi d ti giao im ca d v (P ). Vit phng trnhng thng .

4. Hnh ChiuA. Kin Thc Cn Nh

1. Hnh chiu ca im M trn mt phng (). Vit phng trnh ng thng qua M v vung gc vi (). Hnh chiu H l giao im ca v ().

Lu . Mt phng () ct mt cu (S) theo giao tuyn l ng trn c tm l hnh chiu ca I trn ().

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Chuyn 6. Phng Php Ta Trong Khng Gian

2. Hnh chiu ca im M trn ng thng d. Ly H d. Tnh MH. H l hnh chiu ca M trn d MH.ud = 0.

3. Hnh chiu ca ng thng d trn mt phng ().TH1: d ct (). Tm giao im a ca d v (). Ly M c th trn d. Tm hnh chiu M ca M trn d. Hnh chiu d l ng thng AM .TH2: d song song (). Ly M c th trn d. Tm hnh chiu M ca M trn d. Hnh chiu d qua M v song song vi d.

B. Bi Tp

6.61. Trong khng gian Oxyz, cho im A (1; 4; 2) v mt phng () : x+ y+ z 1 = 0. Tm to im H l hnhchiu vung gc ca A ln (). Tm to im A i xng vi A qua ().

6.62. Trong khng gian Oxyz, cho A (1; 0; 0) v :x 2

1=y 1

2=z

1. Tm to im H l hnh chiu vung

gc ca A ln . Tm to im A i xng vi A qua .

6.63. (D-06) Trong khng gian Oxyz, cho im A (1; 2; 3); d1 :x 2

2=y + 2

1 =z 3

1v d2 :

x 11 =

y 12

=

z + 1

1. Tm A i xng vi A qua d1. Vit phng trnh ng thng qua A, vung gc d1 v ct d2.

6.64. Trong khng gian Oxyz, cho d :x

2=y 11 =

z 31

v (P ) : x+ y+ z 10 = 0. Vit phng trnh hnh chiud ca d ln (P ).

6.65. Trong khng gian Oxyz, cho d :x 1

2=y + 1

1=z 2

2v (P ) : x+ 2y 2z 1 = 0.Vit phng trnh hnh

chiu d ca d ln (P ).

6.66. Trong khng gianOxyz, lp phng trnh d i xng ca d :x 4

3=y 1

1=z 15 qua (P ) : xy+2z3 = 0.

6.67. (A-08) Trong khng gian Oxyz, cho im A (2; 5; 3) v ng thng d :x 1

2=y