Bai Tap Toan Lop 11(Full)

8/10/2019 Bai Tap Toan Lop 11(Full)

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Trng THPT Ng Thi Nhim Bi tp ton 11

64

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2


63

2

Bi 1: Tm

a)6

293lim

3

23

2

+ xx

xxx

x b)

21

3 2lim

1x

x

x

+

Bi 2: Xt tnh lin tc ca hm ssau trn tp xc nh ca n: + +

= +

23 2

, khi x 2( ) 2

3 , khi x = -2

x x

f x x

Bi 3: Cho hm sy = f(x) = 2x3 6x +1 (1)a) Tm o hm cp hai ca hm s(1) ri suy ra ( 5)f .b)

Vit phng trnh tip tuyn ca th hm s (1) tiim Mo(0; 1).

c) Chng minh PT f(x) = 0 c t nht mt nghim nmtrong khong (-1; 1).

Bi 4: Cho hnh chp S. ABCD c y ABCD l hnh thoi cnh

a c gc BAD = 600

v SA=SB = SD = a.a)

Chng minh (SAC) vung gc vi (ABCD).b) Chng minh tam gic SAC vung.c) Tnh khong cch tS n (ABCD).

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MT STHI THAM KHO

1

Cu 1: Tnh gii hn ca hm s

a)

2

3

2 9 9lim

3xx x

x

b)

22 4 1lim

3 2xx x

x

+

+

Cu 2: Xt tnh lin tc ca hm strn tp xc nh ca n:

f(x) =

22 10 22 4

4 17 2

x xx

x

x x

+ + < +

+

nu

nu

Cu 3: Tnh o hm ca cc hm s:a) y = 3x3- 4x2+ 8

b) y =22 5 1

3 4

x x

x

+

c) y = 3sin3x - 3cos24xCu 4:

a) Vit phng trnh tip tuyn ca thhm s(C)y = - 2x4+ x2 3 ti im thuc (C) c honh x0= 1.

b) Cho hm sy = x.cosx.Chng minh rng: x.y 2(y - cosx) + x.y = 0

Cu 5: Cho hnh chp S.ABC c y l tam gic cn B vABC=1200, SA (ABC) v SA = AB = 2a. Gi O l trungim ca on AC, H l hnh chiu ca O trn SC.

a) Chng minh: OB SC.b) Chng minh: (HBO) (SBC).c) Gi D l im i xng vi B qua O. Tnh khong

cch gia hai ng thng AD v SB.


3

Chng I:HM SLNG GIC PHNG TRNH

LNG GIC

PHN 1. HM SLNG GIC

Bi 1.Tm tp xc nh ca cc hm ssau:

1.1

sin1

+=

xy

x 2.

3sin2

2cos3=

xy

x

3. cot(2 )4

= y x 4. 2tan( 5 )

3= +y x

5.1

cos1

=

+

xy

x 6.

sin 2

cos 1

+=

+

xy

7.1

sin cos=

y

x x 8.

2 2

3 tan

cos sin

+=

xy

x x

9. sin coscos 1 1 sin

= + +

x xyx x

10. 212 sin tan 1= + y x

x

Bi 2.Xc nh tnh chn, lca cc hm s:

1.cos3x

yx

= 2. 2 2siny x x=

3. 2siny x x= + 4. 21

tan 12

y x= +

5. 23sin cosy x x= 6. tan 2 cosy x x= + Bi 3.Tm gi trln nht, gi trnhnht ca cc hm s:

1. y 2sin(x ) 33

= + 2.1

y=3- cos2x2

3.21 3cos

y=2

x+ 4. 2 4sin cosy x x=

5.2

4sin cos2y x x= 6. 3 cos2 1y x= +

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4

7. 7 3 sin3y x= 8. 2 25 2sin cosy x x=

Bi 4.Hy xt sbin thin v vthcc hm ssau:1. siny x= 2. 2 siny x=

3. sin( )3

y x

= + 4. cos 1y x= +

PHN 2. PHNG TRNH LNG GIC

DNG 1. PHNG TRNH LNG GIC CBN

Bi 1. Gii cc phng trnh sau:

1.1

sin32

x= 2.2

cos22

x=

3. tan( ) 34

x

= 4. s in2 s in2 cos 0x x x =

5. s in3 cos2 0x x = 6. t an4 cot 2 1x x=

7. 2 cos( ) 1 06

x

+ = 8. tan(2 ) t an3 03

x x

+ + =

9. 2cos 2sin 02

xx = 10. 4 4

2cos sin

2x x =

11.1

sin cos sin cos2 3 3 2 2

x x + =

12. 3 32

sin cos cos sin8

x x x x =

13. 2 2 2cos cos 2 cos 3 1x x x+ + =

14.2 2 17sin 2 cos 8 sin( 10 )

2x x x

= +

15.

4 6

cos sin cos2x x x+ =


61

3. Dng v tnh di on vung gc chung ca AB v

SD

4.

Tnh : d[ ])(, SACM

Bi 6. Cho hnh lng trABC.ABCc AA(ABC) v AA

= a, y ABC l tam gic vung ti A c BC = 2a, AB = a 3 .1. Tnh khong cch tAAn mt phng (BCCB).2. Tnh khong cch tA n (ABC).3. Chng minh rng AB (ACCA) v tnh khong cch

tAn mt phng (ABC).

Bi 7. Cho hnh lp phng ABCD.ABCD.

1.

Chng minh: BD (BAC); BD (ACD)

2. Tnh d (BA'C'),(ACD')

3.

Tnh d (BC'),(CD')

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1. OA v BC 2. AI v OC.Bi 2. Cho hnh chp SABCD, y ABCD l hnh vung tm O,cnh a, SA (ABCD) v SA = a. Tnh khong cch gia haing thng:

1. SC v BD. 2. AC v SD.

Bi 3. Cho hnh chp S.ABCD c y ABCD l hnh vung

canh a, SA (ABCD) v SA = 3a . Tnh:

1. Gia SC v BD ; gia AC v SD.

2.

d [ ])(, ABCDA

3.

d [ ])(, SBCO vi O l tm ca hnh vung.

4.

d [ ])(, ABCDI vi I l trung im ca SC.

Bi 4. Cho hnh chp S.ABCD c y ABCD l hnh thang

vung ti A v D AB = DC = a , SA (ABCD) v SA = 2a

Tnh :

1. d [ ])(, SCDA ; d [ ])(, SBCA

2. d [ ])(, SCDAB

3. d [ ])(, SCDAB

4.

d [ ])(, SBCDE , E l trung im ca ABBi 5. Cho hnh chp S.ABCD c y l hnh vung cnh a ,tam

giac SAD u v (SAD) (ABCD) .gi I l trung im ca Sb

va K =CM BI

1. Chng minh (CMF) (SIB)

2.

Chng minh : tam giac BKF cn ti K


5

16.1 cos4 s in4

02s in2 1 cos4

x x

x x

=

+

17. 2 2 1sin cos cos2

x x x ++ =

18.

2(2 3)cos 2sin ( )2 4 1

2 cos 1

xx

x

=

Bi 2.Gii v bin lun phng trnh:1. sin 2 1x m=

2. (4 1)cos cos 8m x m x = 3. 4tan ( 1) tanx m m x = +

4. 2(3 2)cos2 4 sin 0m x m x m + + = Bi 3.Tm m phng trnh:

1. 2 sin( )4

x m

+ = c nghim (0; )2

x

2.7

(2 )sin( ) (3 2)cos(2 ) 2 02m x m x m

+ + + + = cnghim.

DNG 2. PHNG TRNH BC HAI I VI MTHM SLNG GIC

Bi 1.Gii cc phng trnh sau:

1. 24cos 2( 3 1)cos 3 0x x + + = 2. 22cos x 5sinx 4 0+ = 3. 2cos2x 8cosx 5 0+ = 4. 2cosx.cos2x 1 cos2x cos3x= + +

5. 22

33 2 tan

cos = + xx

6. 5tan x 2cotx 3 0 =

7. 26sin 3 cos12 4x x+ =

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6

8. 2cos2 3cos 4 cos2

x xx

=

9. 2cos4cot tansin2

xx xx

= +

10.2cos (2sin 3 2) 2sin 3

11 sin2

x x x

x

+ + =

+

11. 4 43tan 2 tan 1 0x x+ =

12.1 1

cos sinsin cos

x xx x

=

13. 22

1 1cos 2(cos ) 1

coscosx x

xx+ + =

14.2 2

1 14

sin cossin cos x xx x+ =

Bi 2.Tm m phng trnh sau c nghim:

1. 2cos (1 )cos 2 6 0x m x m+ + =

2. 24 cos 2 4 cos2 3 3 0x x m =

Bi 3.Cho phng trnh: cos2 ( 2)sin 1 0x a x a+ + = 1. Gii phng trnh cho khi a = 1.

2. Vi gi trno ca a th phng trnh cho cnghim?

DNG 3. PHNG TRNH BC NHT THEOSINu V COSu

Bi 1.Gii cc phng trnh sau:

1. 2sincos3 = xx

2. 1sin3cos = xx


59

1. Chng minh: (SAB) (SAD), (SAB) (SBC).

2. Tnh gc gia hai mp (SAD), (SBC).

3.

Gi H, I ln lt l trung im ca AB v BC. Chng

minh: (SHC) (SDI).

Bi 10. Cho tam gic ABC vung ti A. Gi O, I, J ln lt l

trung im ca BC v AB, AC. T O k on thng

OS (ABC).

1.

Chng minh: (SBC) (ABC).2. Chng minh: (SOI) (SAB).

3. Chng minh: (SOI) (SOJ).

Bi 11. Cho tam din ba gc vung Oxyz (3 tia Ox, Oy, Oz i

mt vung gc). Ln lt ly trn Ox, Oy, Oz cc im B, C, A

sao cho OA = a, OB = b, OC = c. Cc ng cao CH va BK catam gic ABC ct nhau ti I.

1. Chng minh: (ABC) (OHC).

2. Chng minh: (ABC) (OKB).

3. Chng minh: OI (ABC).

4. Gi , , ln lt l gc to bi OA, OB, OC vi OI.

Chng minh: cos2+ cos2+ cos2= 1.

KHONG CCH

Bi 1. Cho hnh tdin OABC, trong OA, OB, OC = a. Gi I

l trung im ca BC. Hy dng v tnh di on vung gcchung ca cc cp ng thng:

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1. Chng minh: (SBC) (ABC).

2. Chng minh: (SOI) (ABC).

Bi 6. Cho hnh chp S.ABCD, y l hnh vung cnh a. Tam

gic SAB u nm trong mt phng vung gc vi y. I, J, K

ln lt l trung im ca AB, CD, BC.

1. Chng minh: SI (ABCD).

2. Chng minh: trn mt phng SAD v SBC l nhng tam

gic vung.3.

Chng minh: (SAD) (SAB), (SBC) (SAB).

4.

Chng minh: (SDK) (SIC).

Bi 7. Cho tdin ABCD c cnh AD (BCD). Gi AE, BF

l hai ng cao ca tam gic ABC, H v K ln lt l trc tm

ca tam gic ABC v tam gic BCD.1. Chng minh: (ADE) (ABC).

2.

Chng minh: (BFK) (ABC).

3.

Chng minh: HK (ABC).

Bi 8.Trong mp (P) cho hnh thoi ABCD vi AB = a, AC =

2 63

a . Trn ng thng vung gc vi mp (P) ti giao im O

ca hai ng cho hnh thoi ta ly S sao cho SB = a.

1. Chng minh: SAC vung.

2.

Chng minh: (SAB) (SAD).

Bi 9. Cho hnh vung ABCD. Gi S l im trong khng gian

sao cho SAB l tam gic u v (SAB) (ABCD).


7

3. s in3 3 cos3 2x x+ =

4. 22 cos 3 s in2 2x x =

5. 2s in2 cos2 3 cos 4 2 0x x x+ + =

6. )7sin5(cos35sin7cos xxxx =

7.4

1)

4(cossin 44 =++

xx

8. tan 3cot 4(sin 3 cos )x x x x = +

9.2 1

sin 2 sin 2x x+ =

10. 33sin3 3 cos9 1 4sin 3x x x = +

11.3(1 cos 2 )

cos2sin

xx

x

=

12.cos sin

cot tansin cos

x xx x

x x

=

Bi 2. nh m phng trnh sau y c nghim:1. sin 2 cos 3m x x+ = 2. s in2 cos2 2 0x m x m+ + = 3. cos3 ( 2)s in3 2m x m x+ + = 4. (sin 2cos 3) 1 cosx x m x+ + = + 5. (cos sin 1) sinm x x x = 6. (3 4 )cos2 (4 3)s in2 13 0m x m x m+ + + =

Bi 3. Cho phng trnh: sin cos 1x m x+ = 1. Gii phng trnh khi 3m= .2. nh m phng trnh trn v nghim.

DNG 4. PHNG TRNH THUN NHT BC HAITHEO SINu V COSu

Bi 1.Gii cc phng trnh sau:1. 2 2sin x 3sinxcosx 4cos x 0+ =

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8

2. 2 23sin x 8sinxcosx ( 8 3 9)cos x 0+ + =

3. 2 24sin x 3

sin2x 2cos x 4+ =

4. 2 22sin x 5sinx.cosx cos x 2=

5. 2 24sin 3 3 sin 2 cos 42 2

x xx+ =

6. 2 22sin 6sin cos 2(1 3)cos 5 3x x x x+ + + = +

7. 3 2 3sin 2sin cos 3cos 0x x x x+ =

8. 3 2 34sin 3sin cos sin cos 0x x x x x+ =

9. 3 3 2 2sin 3 cos sin cos 3 sin cosx x x x x x =

10.2

2 tan cot 3sin2

x xx

+ = +

Bi 2.Tm m phng trnh sau c nghim:

1. 2 2sin 2s in2 3 cos 2m x x m x+ + =

2. 2 2sin s in2 ( 1)cos 0x m x m x + =

DNG 5. PHNG TRNH I XNG PHN XNG

Bi 1.Gii cc phng trnh sau:1. 2(sin cos ) 3sin cos 2 0x x x x+ + + =

2. ( )3 sinx cosx 2sin2x 3 0+ + + =

3. ( )sin2x 12 sinx cosx 12=

4. ( )2 cosx sinx 4sinxcosx 1+ = + 5. cosx sinx 2sin2x 1 0=

6. (1 2)(sin cos ) 2sin cos 1 2 0x x x x+ + =

7. 3 3sin cos 1 sin cosx x x x+ =

8. 3 3sin cos 2(sin cos ) 1x x x x+ = +

9. tan cot 2(sin cos )x x x x+ = +


57

3. Gi BE, DF l hai ng cao ca tam gic SBD. Chng

minh rng: (ACF) (SBC), (AEF) (SAC).

Bi 2. Cho tdin ABCD c cc mt ABD v ACD cng vung

gc vi mt BCD. Gi DE ,BK l ng cao tam gic BCD v

BF l ng cao tam gic ABC

1. Chng minh : AD (BCD)

2. Chng minh : (ADE) (ABC)

3.

Chng minh : (BKF) (ABC)4. Chng minh : (ACD) (BKF)

5. Gi O v H ln lt l trc tm ca hai tam gic BCD v

ABC chng minh : OH (ABC)

Bi 3. Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh

a. SA= SB= SC=a. Chng minh :1. (ABCD) (SBD)

2. Tam gic SBD l tam gic vung.

Bi 4. Cho tam gic u ABC cnh a, I l trung im ca cnh

BC, D l im i xng ca A qua I. Dng on SD =6

2

a

vung gc vi (ABC). Chng minh:

1. (SAB) (SAC).

2.

(SBC) (SAD).

Bi 5. Cho hnh chp S.ABCD c y ABC l tam gic l tam

gic vung ti A, AB = 2a, AC = a, SA = SB = SC = 2a . Gi

O l trung im ca BC, I l trung im ca AB.

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3. Tnh gc [(SMC), (ABC)].

Bi 7. Cho hnh chp S.ABCD c y ABCD l hnh thang

vung ti A v D vi AB = 2a, AD = DC = a, SA = 2a . SA

(ABCD). Tnh gc gia cc mt phng.

1. (SBC) v (ABC).

2.

(SAB) v (SCB).

3. (SCB) v (SCD).

Bi 8. Cho hnh chp S.ABCD c y l hnh thoi ABCD tm

O, cnh a ABC= 600, SO (ABCD) v SO =3

4

a. Tnh so

nhdin cnh AB.


cnh a, tm O, SA

(ABCD) v SA = x (x>0).1.

Tnh s[S, BC, A] theo a v x. Tnh x theo a so nh

din trn bng 600.

2.

Tnh s[B, BC, D] theo a v x. Tnh x theo a so nh

din trn bng 1200

HAI MT PHNG VUNG GC

Bi 1. Cho hnh chp S.ABCD, y ABCD l hnh vung, SA

(ABCD).

1.

Chng minh: (SAC) (SBD).

2. Chng minh: (SAD) (SCD), (SAB) (SBC).


9

10.cos2

sin cos1 s in2

xx x

x+ =

Bi 2.nh m phng trnh sau c nghim:1. sin cos 1 s in2x x m x+ = +

2. 2s in2 2 2 (sin cos ) 1 6 0x m x x m + + =

DNG 6. PHNG TRNH LNG GIC KHNG MUMC

Bi tp. Gii cc phng trnh sau:1. sin .s in2 1x x=

2. 2 1007cos 8sin 8x x+ =

3. sin cos 2(2 s in3 )x x x+ =

4. 3 3 4sin cos 2 s inx x x+ =

MT STHI I HC1. 2(1 2sin ) cos 1 sin cosx x x x+ = + +

2. 3 cos5 2sin 3 cos 2 sin 0x x x x =

3. 3sin cos sin 2 3 cos3 2(cos 4 sin )x x x x x x+ + = +

4.(1 2sin ) osx

3

(1 2sin )(1 s inx)

x c

x

=

+

5. sin 3 3 cos3 2sin 2x x x = 6. 2sin (1 cos 2 ) sin 2 1 2cosx x x x+ + = +

7. 3 3 2 2sin 3 cos sin cos 3 sin cosx x x x x x =

8.1 1 7

4sin( )3sin 4sin( )2

xx

x

+ =

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10

9. 2(sin cos ) 3 cos 22 2

x xx+ + =

10.

2

2sin 2 sin 7 1 sinx x x+ =

11. 2 2(1 sin )cos (1 cos )sin 1 sin 2x x x x x+ + + = + 12. cos3 cos 2 cos 1 0x x x+ =

13. cot sin (1 tan tan ) 42

xx x x+ + =

14.6 62(cos sin ) sin cos

02 2sin

x x x x

x

+ =

15. 4 4 3cos sin cos( )sin(3 ) 04 4 2

+ + =x x x x

16. 1 sin cos sin 2 cos 2 0x x x x+ + + + = 17. 2 2cos 3 cos 2 cos 0x x x = 18. 25sin 2 3(1 sin ) tanx x x = 19. (2cos 1)(2sin cos ) sin 2 sinx x x x x + =

20.

2

cot tan 4sin 2 sin2x x x x + =


55

Bi 4. Cho hnh vung ABCD v tam gic u SAB cnh a nm

trong hai mt phng vung gc nhau. Gi I l trung im ca

AB.

1. Chng minh: SI (ABCD) v tnh gc gia SC v

(ABCD).

2. Gi J l trung im CD. Chng t: (SIJ) (ABCD) . Tnh

gc hp bi SI v (SDC).

Bi 5. Cho hnh chp S.ABCD c y ABCD l hnh vung tm

O, cnh a, SA (ABCD) v SA = a. Tnh:

1.

[SAB, (SCD)].

2. [SAB, (SBC)].

3. [SAB, (SAC)].

4.

[SCD, (ABCD)].5. [SBC, (SCD)].

6. s[S, BC, A].

7. s[C, SA, D].

8.

s[A, SB, D].

9.

s[B, SC, A].Bi 6. Cho hnh chp S.ABCD c y ABC l tam gic vung

ti B, AB = 2a, BC = 3a , SA (ABC) v SA = 2a. Gi M l

trung im ca AB.

1. Tnh gc [(SBC), (ABC)].

2.

Tnh ng cao AK ca AMC.

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4. Gi d l ng thng vung gc vi (ABC) ti trung im

K ca BC tm d ( ).

- GC GIA NG THNG V MT PHNG

- GC GIA HAI MT PHNG


cnh a, tm O, SO (ABCD), M, N ln lt l trung im ca

SA v BC, bit 0( ,( )) 60MN ABCD = .

1. Tnh MN v SO.

2. Tnh gc gia MN v mp(BCD).


cnh a. SA (ABCD) v SA = a 6 . Tnh gc gia:

1. SC v (ABCD)

2.

SC v (SAB)

3. SC v (SBD)

4.

SB v (SAC)

Bi 3. Cho t din ABCD c AB (BCD) v AB = 3a ,

BCD l tam gic u cnh a. Tnh gc gia:

1. AC v (BCD).

2. AD v (BCD).

3. AD v (ABC).


11

Chng II. T HP XC SUT

PHN 1. HON VN- CHNH HP - THP

Bi 1.C 25 i bng tham gia thi u, c2 i th vi nhau2 trn ( i v v). Hi c tt cbao nhiu trn u?Bi 2.

1. Tcc chs1, 2, 3, 4, 5 c thlp c bao nhiu stnhin c 5 chs?

2.

Tcc chs0, 1, 2, 3, 4, 5, 6 c thlp c baonhiu stnhin c 3 chsv l schn?3. C bao nhiu stnhin c 6 chsi mt khc nhauv chia ht cho 5?

Bi 3.Mt hi ng nhn dn c 15 ngi, cn bu ra 1 chtch, 1 ph chtch, 1 thk. Hi c my cch nu khng aic kim nhim?Bi 4.Trong mt tun, An nh mi ti i thm 1 ngi bn

trong s10 ngi bn ca mnh. Hi An c thlp c baonhiu khoch thm bn nu:1.

C ththm 1 bn nhiu ln?2. Khng n thm 1 bn qu 1 ln?

Bi 5. C bao nhiu cch xp 10 hc sinh thnh mt hng dc?Bi 6. C bao nhiu cch xp 5 bn A, B,C,D,E vo mt ghdi5 chnu:

1. Bn C ngi chnh gia.

2.

Hai bn A v E ngi hai u gh.Bi 7.Tcc chs1,2,3,4,5,6 c ththit lp c bao nhiusc 6 chskhc nhau m hai chs1 v 6 khng ng cnhnhau?Bi 8.C 2 sch Ton khc nhau, 3 sch L khc nhau v 4sch Ha khc nhau.Cn sp xp cc sch thnh mt hng saocho cc sch cng mn knhau. Hi c bao nhiu cch?Bi 9.Gii :

1.

P2.x2 P3.x = 8

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12

2.

1

1

1

6

x x

x

P P

P

+

=

3.

12

4 15. +

+ ++

++

+ nnn Bi 4.Chng minh vi mi stnhin 2n , ta c cc btng thc sau:

1. 133 +> nn

2.2

32 > nn

3. 322 1 +>+ nn

Bi 5.Chng minh vi mi stnhin 3n , ta c:122 +> nn

DY S

Bi 1. Xt tnh n iu cc dy ssau :

1. 21 1nu

n= + 2. 12 3+= n

n

nu

3.1

2

n

nu =

4. nnun += 1 .

5.2 1

2

n

n nu

= 6.

nn

nu

2

2+=

7. nu nn

= 3 8. 12 = nnun

.

Bi 2.Xt tnh bchn cc dy ssau :

1. 23 = nun 2.1

( 1)nu

n n=

+

3. 13.2nnu = 4. nnu )3(=

5.34

34

+

=

n

nun 6.

2

1

1n

nu

n

=

+

Trng THPT N g Thi N him Bi tp ton 11

47

2. GisAB CD th MN QG l hnh g? Tnh SMN PQbitAM = x, AB = AC = CD = a. Tnh x din tch ny lnnht.

HAI MT PHNG SONG SONG

Bi 1. Cho hai hnh bnh hnh ABCD , ABEF c chung cnhAB v khng ng phng . I, J, K ln lt l trung im ca cccnh AB, CD, EF. Chng minh:

1. (ADF) // (BCE).

2.

(DIK) // (JBE).Bi 2.Cho tdin ABCD.Gi H, K, L l trng tm ca cc tamgic ABC, ABD, ACD. Chng minh rng (HKL)//(BCD).Bi 3. Cho hnh chp S.ABCD y l hnh bnh hnh tm O.Tam gic SBD l tam gic u. Mt mp () di ng song songvi (SBD) qua im I trn on AC. Xc nh thit din cahnh chp ct bi ().Bi 4.Cho hnh chp S.ABCD c ABCD l hnh thang vung

ti A v D; AD = CD = a ; AB = 2a, tam gic SAB vung cntiA.Trn cnh AD ly im M. t AM =x. Mt phng () quaM v //(SAB).

1.

Dng thit din ca hnh chp vi ().2. Tnh din tch v chu vi thit din theo a v x.

Bi 5.Cho hai mp (P) v (Q) song song vi nhau v ABCD lmt hnh bnh hnh nm trong mp (P). cc ng thng song

song i qua A, B, C, D ln lt ct mp (Q) ti cc im A', B',C', D'.1. Tgic A'B'C'D' l hnh g?2. Chng minh (AB'D') // (C'BD).3.

Chng minh rng on thng A'C i qua trng tm ca haitam gic AB'D' v C'BD. Hai mp (ABD), (CBD) chiaon A'C lm ba phn bng nhau.

HNH LNG TR

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46

Chng minh : MN // (BCD) v MN // (ABC).Bi 2.Cho tdin ABCD .Gi I, J l trung im ca BC v CD

1.

Chng minh rng BD//(AIJ)

2.

Gi H, K l trng tm ca cc tam gic ABC v ACD.Chng minh rng HK//(ABD)

Bi 3.Cho hnh chp S.ABCD c ABCD l hnh bnh hnh .Gl trng tm ca tam gic SAB v E l im trn cnh AD saocho DE = 2EA. Chng minh rng GE // (SCD).Bi 4.Hnh chp S.ABCD c y ABCD l hnh bnh hnh.Gi M , N theo thtl trung im ca cc cnh AB, CD .

1. Chng minh MN // (SBC) v MN // (SAD)2.

Gi P l trung im ca cnh SA. Chng minh SB //(MN P) v SC // (MN P).

Bi 5.Cho hnh chp S.ABCD. M, N l hai im bt k trnSB v CD. () l mt phng qua MN v song song vi SC.

1. Tm cc giao tuyn ca ( ) vi cc mt phng (SBC),(SCD) v (SAC).

2.

Xc nh thit din ca S.ABCD vi mt phng () .

Bi 6.Cho hnh chp S.ABCD c ABCD l hnh bnh hnh .GiM,N l trung im SA,SB. im P thay i trn cnh BC

1. Chng minh rng CD//(MN P)2.

Dng thit din ca hnh chp vi mt phng (MN P) .Chng minh rng thit din l 1 hnh thang.

3. Gi I l giao im 2 cnh bn ca thit din ,tm qutchim I

Bi 7. Cho hnh chp S.ABCD. M, N l hai im trn AB,CD, () l mt phng qua MN v song song vi SA.

1. Xc nh thit din ca hnh chp v mt phng ().2. Tm iu kin ca MN thit din l hnh thang.

Bi 8. Cho tdin ABCD. Tim M trn AC ta dng mt mp() song song AB v CD. Mp ny ln lt ct BC, BD, AD tiN , P, Q.

1.

Tgic MN QG l hnh g?


19

Bi 3. Cho dy s ( )nu xc nh bi:

+

+=

=

+

1

2

1

1

1

n

n

n

u

uu

u

; 1n .

Chng minh rng nu bchn trn bi 2

3v bchn di bi 1.

Bi 4.Cho dy s ( )nu xc nh bi:

+=

=

+ 2

1

2

1

1

n

n

uu

u

; 1n .

Chng minh rng nu l dy gim v bchn.

Bi 5.Cho dy s ( )nu xc nh bi:

++=

=

+n

nn nuu

u

2).1(

1

1

1

; 1n .Chng minh rng :

1. ( )nu l dy tng.

2. nn nu 2).1(1 += , 1n .

CP SCNG

Bi 1.Tm shng u v cng sai ca cc cp scng, bit :

1.

=+

=+

17

10

61

531

uu

uuu 2.

=

=

75

8

152

37

uu

uu

3.

=

=+

129

14

12

53

s

uu

4.

=+

=+

1170

60212

24

157

uu

uu

5.

=

=++

24

25

82

541

uu

uuu 6.

=

=

75.

8

72

37

uu

uu

Bi 2.1. Cho cp scng c 1a =10, d = -4 .Tnh 10a v 10S .

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44

2. Tm giao im ca SD vi mt phng (AMN ) ?3.

Tm tit din to bi mt phng (AMN ) vi hnh chpBi 18: Hnh chp SABCD c y ABCD l hnh bnh hnh .

M l trung im SC1. Tm giao im I ca AM vi (SBD) ? Chng minh IA= 2IM .

2. Tm giao im F ca SD vi (AMB) ? Chng minh F ltrung im SD ?

3.

Xc nh hnh dng tit din to bi (AMB) vi hnhchp.

4. Gi N l mt im trn cnh AB .Tm giao im ca

MN vi (SBD) ?

HAI NG THNG SONG SONG

Bi 1. Cho tdin ABCD. Gi I, J, K, L theo th t l trungim ca cc cnh AB, BC ,CD ,DA Chng minh : IJ//KL vJK//IL .Bi 2. Cho tdin ABCD .Gi H, K l trng tm ca cc tamgic BCD v ACD .Chng minh rng HK//AB.Bi 3.Cho hnh chp S.ABCD c y l mt tgic li. Gi M,N ,E ,F ln lt l trung im ca cc cnh bn SC, SB, SC vSD.

1. Chng minh rng ME//AC , N F//BD2. Chng minh rng ba ng thng ME ,N F ,v SO(O l

giao im ca AC v BD) ng qui

3.

Chng minh rng 4 im M,N

,E,F ng phngBi 4.Cho hnh chp S.ABCD c ABCD l hnh bnh hnh. GiH, K l trung im SA, SB.

1. Chng minh rng HK//CD2.

Trn cnh SC ly im M. Dng thit din ca hnh chpvi mt phng (MKH).

Bi 5.Cho hnh chp S.ABCD y l hnh thang vi cc cnhy l AB v CD. Gi I, J lm lt l trung im ca DA v BC

v G l trng tm tam gic SAB.1. Tm giao tuyn ca (SAB) v (IJG)


21

Bi 11. Chng minh rng ba sdng a, b, c lp thnh cp s

cng khi v chkhi cc s:baaccb +++

1,

1,

1lp

thnh cp scng.Bi 12.Tm bn shng lin tip ca mt cp scng bit tngca chng l 20 v tch ca chng l 348.

CP SNHN

Bi 1.Trong cc cp snhn di y, hy tnh shng nu

chra:1. 2; 1;

2

1;

4

1; ?7 =u

2. -3; 6; -12; 24; ?10 =u

3. 1;3

1;

9

1;

27

1; ?8 =u

Bi 2. Tm shng u, cng bi ca cc cp snhn, bit :

1.

=

=

192

96

6

5

u

u 2.

=+

=++

10

21

42

531

uu

uuu

3.

=

=+

240

90

62

53

uu

uu 4.

=

=

144

72

35

24

uu

uu

5.

=+

=+

325

65

71

531

uu

uuu 6.

=+

=+

20

10

653

542

uuu

uuu.

Bi 3.Tm cp snhn ( nu ) bit:1 2 3 4

2 2 2 21 2 3 4

15

85

u u u u

u u u u

+ + + =

+ + + =

Bi 4. Hy tm shng ca cp snhn, bit cp snhn :1.C 5 shng vi cng bi dng, shng thhai bng 3

v shng thtbng 6.

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22

2. C 5 shng vi cng bi bng1

4shng thnht v

tng ca hai shng du bng 24.

Bi 5.Cho mt cp snhn c 7 shng, shng thtbng 6v shng thby gp 243 ln shng thhai. Hy tmcc shng cn li ca cp snhn .

Bi 6. Hy tm shng tng qut ca cp snhn ( nu ) c

=+

=+

123

16

43

52

uu

uu.

Bi 7.Tnh tng:

1. ...3

2.)1(...

9

4

3

21 1 +

+++= +

n

nS

2. ...1 32 +++= aaS vi21

1

+=a

Bi 8.Tnh tng tt ccc shng ca cp snhn (un) bit:1

2

2

2

64 2n

u

u

u

=

=

=

Bi 9.Mt cp scng v mt cp snhn u l cc dy tng.Cc shng thnht u bng 3, cc shng thhai bng

nhau. Tsgia cc shng thba ca cp snhn v cp scng l 9/5 .Tm hai cp sy.Bi 10. Tm hai sa, b bit rng 1,a,b l cp scng v 1,a2,b2l cp snhn.


43

Bi 13: Cho hnh chp S.ABCD y l hnh thang, cnh y lnAB. Gi I, J, K ln lt l cc im nm trn SA, AB, CD

1.

Tm giao im ca IK v (SBD).

2.

Tm giao im ca SD v (IJK).3. Tm giao im ca SC v (IJK) .

THIT DIN

Bi 1: Cho tdin ABCD. Gi M, N ln lt l trung im cccnh AB v CD. P l im nm trn cnh AD nhng khng ltrung im. Tm thit din ca tdin ct bi mt phng(MN P).Bi 2: Cho tdin ABCD. Trn cc on AC, BC, BD ly ccim M, N , P sao cho MN khng song song vi AB, N P khngsong song vi CD. Xc nh thit din to bi mt phng(MN P) v tdin ABCD.Bi 6: Cho hnh chp SABCD. Gi M l 1 im thuc mintrong ca tam gic SCD.

1.

Tm giao tuyn ca hai mt phng (SBM) v (SAC).2. Tm giao

im c

a BM v m

t ph

ng (SAC).

3.

Tm thit din ca hnh chp ct bi mt phng (ABM).Bi 9: Cho hnh chp SABCD y l hnh bnh hnh tm O.Mt im M trn cnh SD sao cho SD = 3SM.

1. Tm giao tuyn ca (SAC) v (SBD).2. Xc nh giao im I ca BM v (SAC). Chng t I l

trung im ca SO.3. nh thit din ca hnh chp SABCD v (MAB).

Bi 14: Cho tdin ABCD ; im I nm trn BD v ngoiBD sao cho ID = 3IB; M; N l hai im thuc cnh AD; DC sao

cho MA=2

1 MD; N D =2

1N C

1. Tm giao tuyn PQ ca (IMN ) vi (ABC) ?2. Xc dnh thit din to bi (IMN ) vi tdin ?3.

Chng minh MN ; PQ ; AC ng qui ?Bi 17: Hnh chp SABCD c y ABCD l hnh thang vi AB

l y . Gi M ; N l trung im SB ; SC .1. Tm giao tuyn ca (SAD) v (SBC) ?

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42

Bi 6: Cho hnh chp SABCD c ABCD l hnh thang vi yln l AB. Gi I, J ln lt l trung im ca SA, SB. M l imtu trn cnh SD.

1.

Tm giao tuyn ca(SAD) v (SBC).2. Tm giao im K ca IM vi mt phng (SBC).3.

Tm giao im N ca SC vi mt phng (IJM).Bi 7: Cho hnh chp SABCD c y l hnh bnh hnh. Gi Ml trung im ca SC.

1.

Tm giao im I ca ng thng AM vi mt phng(SBD).

2. Chng minh IA= 2IM.

3.

Tm giao im F ca SD v (ABM).4. im N thuc AB. Tm giao im ca MN v (SBD).

Bi 8: Cho tgic ABCD nm trong mt phng (P) c hai cnhAB v CD khng song song. Gi S l im nm ngoi (P) v Ml trung im ca on SC.

1. Tm giao im N ca SD v (MAB)2. Gi O l giao im ca AC v BD . CMR: SO, AM, BN

ng quiBi 9: Cho tdin ABCD. Hai im M, N ln lt nm trongtam gic ABC v tam gic ABD. I l im tu trn CD. Tmgiao ca (ABI) v ng thng MN .Bi 10: Cho hnh chp SABCD. Gi I, J l hai im trn cnhAD, SB

1. Tm cc giao im K, L ca IJ v DJ vi (SAC)2. AD ct BC ti O; OJ ct SC ti M. Chng minh A, K, L,

M thng hngBi 11: Cho tdin ABCD. Gi M, N ln lt l trung im caAC, BC. K l im trn cnh BD v khng trng vi trung imca BD.

1.

Tm giao im ca CD v (MN K).2. Tm giao im ca AD v (MN K)

Bi 12: Cho tdin ABCD. M, N l 2 im trn cnh AC, AD.O l 1 im bn trong BCD. Tm giao im ca:

1.

MN v (ABO).2. AO v (BMN ).


23

CHNG IV. GII HN

GII HN CA DY S

Bi 1. Tnh cc gii hn sau:

1. lim92

142

2

+

n

nn 2. lim

37

76

652

+

+

nn

nnn

3. limnn

n

108

25 +

+ 4.

36

4325

4

+

+

nn

nn

5. lim 23

4

11100

3373

nn

nn

+

6. lim )32(3

)31(23

22

nn

nn

+

7. lim23

32

)42(

)2()23(

n

nn

8. lim

7

323

432

)5()51(

nn

nn

+

+

Bi 2.Tnh cc gii hn sau:

1.1

1lim

+

+

n

n 2.

2lim

3 3

+

+

n

nn

3.32

232lim

2

4

+

+

nn

nn 4.

12

21lim

2

+

+

n

nn

5. lim756

14

33 62

+

+++

nn

nnn 6.

12lim

43

+

++

n

nnn

7.

nnn

nn

+

++4 3

2 1lim 8.

23

11lim

2

+

++

n

nn


1.12

13lim

+n

n

2.n

nn

5.37

5.23lim

+

3.11

5)3(

5)3(lim

++

+

+nn

nn

4. lim52

12

24.5

43++

++

nn

nn

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24


1. lim )nn +12 2. lim( 3n n+ )

3. )nnn ++ 1lim2

4. 12

1lim ++ nn

5. )nnn + 1lim 2 6. )nnn +3 32lim7. ( )3 31lim nn + 8.

++ nnnnlim

GII HN CA HM S


1.2

3lim

3

2

1 +

x

x

x

2.622

35lim

23

2

2 +++

++ xxx

xx

x

3.72

15lim

1 +

x

x

x

4. 32

4

2 2

232lim

+

++ xx

xx

x


1. 3 152lim2

3

+ x

xxx 2. 5

152lim

2

5 +

+ x

xx

x

3.

31 1

3

1

1lim

xxx 4.

253

103lim

2

2

2

+ xx

xx

x

5.xx

xx

x 4

43lim

2

2

4 +

+

6.6)5(

1lim

3

1 +

xx

x

x

7.6

44lim

2

23

2

++

xx

xxx

x

8.6

23lim

2

23

2

++

xx

xxx

x

9.6

293lim

3

23

2

+ xx

xxx

x

10.32

1lim

2

4

1 +

xx

x

x


1. .2

35lim

2

2

+ x

x

x 2.

2

153lim

2

x

x

x

3. 11lim0 + x

x

x 3. x

x

x

5

5lim5


41

Bi 3.Cho hnh chp S.ABCD y ABCD l hnh bnh hnh ; Ol giao im hai ng cho; M ; N ln lt l trung im SA;

SD. Chng minh ba ng thng SO; BN ; CM ng quy.

GIAO IM CA NG THNG V MT PHNG

Bi 1: Cho tdin ABCD. Gi M, N ln lt l trung im caAC v BC. Gi K l mt im trn cnh BD khng phi l trungim. Tm giao im ca:

1.

CD v mt phng (MN K)2. AD v mt phng (MN K)

Bi 2: Cho tdin ABCD. Trn cc cnh AB v Ac ln lt lycc im M, N sao cho MN khng song song vi BC. Gi O lmt im nm trong tam gic BCD.

1. Tm giao im ca MN v (BCD)2.

Tm giao tuyn ca (OMN ) v (BCD)3. Mt phng (OMN ) ct cc ng thng BD v CD ti H

v K. Xc nh cc im H v K.Bi 3: Cho hnh chp SABCD. Gi I, J, K ln lt l cc imtrn cc cnh SA, AB, BC. Gi s ng thng JK ct ccng thng AD, CD ti M, N . Tm giao im ca cc ngthng SD v SC vi mt phng (IJK).Bi 4: Cho tdin ABCD. Gi M, N , P l cc im ln lt trncc cnh AC, BC, BD.

1.

Tm giao im ca CP v (MN

D).2.

Tm giao im ca AP v (MN D).Bi 5: Cho 4 im A, B, C, D khng ng phng. Gi M, N lnlt l trung im ca AC v BC. Trn BD ly im P sao choBP=2PD.

1. Tm giao im ca cc ng thng CD vi mtphng(MN P)

2. Tm giao im ca hai mt phng (MN P) v (ACD).

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40

SE, SB ln lt ti M, N . Mt mt phng (Q) qua BC ct SD v

SA ln lt ti H v R.

1.

Gi I l giao im ca AM v DN , J l giao im ca

BH v ER. CMR bn im S, I, J, G thng hng.

2. Gi sK l giao im ca AN v DM, L l giao im

ca BR v EH. CMR ba im S, K, L thng hng.

Bi 9: Cho A; B; C khng thng hng ngoi mt phng ( ) .

Gi M; N ; P ln lt l giao im AB; BC; AC vi . Chngminh M; N ; P thng hng ?Bi 10: Hnh chp S.ABCD c y ABCD l hnh thang hai yl AD; BC. Gi M; N l trung im AB; CD v G l trng tmSAD. Tm giao tuyn ca :

1. (GMN ) v (SAB)2. (GMN ) v (SCD)3.

Gi giao im ca AB v CD l I; J l giao im ca haigiao tuyn cu a v cu b. Chng minh S; I; J thng hng .

4.

CHNG MINH BA NG THNG NG QUI

Bi 1.Cho hnh thang ABCD (AB// CD) im S nm ngoi mtphng cha ABCD. Gi M, N ln lt l trung im ca SC,SD. Gi I l giao im ca AD v BC, J l giao im ca AN vBM.

1.

CMR : S, I, J thng hng.2.

Gi O l giao im ca AC v BD. CMR : SO, AM, BN ng quy.Bi 2.Cho tdin ABCD. M, N ln lt l trung im BC, BD.

Cc im P v S ln lt thuc AD, AC sao cho1

3AR AD= ;

1

3

AS AC= . CMR : ba ng thng AB, MS, N R ng quy.


25

5.25

34lim

25

+ x

x

x 6.

x

xx

x

11lim

2

0

++

7. ( )xxxx

x++ 121lim

2

0 8. xxx

x 3361lim 21 ++

+

9.1

132lim

21

+ x

xx

x

10.23

2423lim

2

2

1 +

xx

xxx

x


1.x

xx

x

+

55lim

0 2.

x

xxx

x

11lim

2

0

+++

3.x

xx

141lim3

0+

4.

x

xx 3

11lim3

0+

5.11

lim30 + x

x

x 6.

x

x

x

+ 51

53lim

4

7.1

lim2

1

x

xx

x 8.

23

1lim

2

3

1 +

+ x

x

x

9.x

xxx

3

0812lim

10.

157lim

23

1 +

xxx

x

Bi 5.Gii hn mt bn:

1.1

3 2lim

1x

x

x+

2.

34

1lim

2

4

3 ++

++ xx

x

x

3. )(lim1

xfx

bit ( )

>+

=

1;1

1;13

2 xx

xxxf

4. )(lim1

xfx

; )(lim3

xfx

bit ( )


26/32


26

1.x

x

x

5sinlim

0 2.

x

x

x 3

2tanlim

0

3.2

0

cos1lim

x

x

x

4.x

xx

x 30 sin

sintanlim

5.x

x

x 2sin

121lim

0

+

6.30 45

sin.3sin.5sinlim

x

xxx

x

7.30

tansinlim

x

xx

x

8.xx

x

x sin

cos1lim

3

0

9.x

xx

x sin

112lim

3 2

0

++

10.2

2

0

cos1lim

x

xx

x

+


1.32

3

662

13lim

xx

xx

x

++

2.3

2x

x 5lim

x 1+

+

3.2

5 3

2lim

3 4 1x

x

x x+

+ 4.

2

5 1lim

3 4 1x

x

x x

+ +

5.2

3

(2 1) (1 3 )lim

( 1)x

x x

x

+ 6. ( ) ( )

( )50

3020

12

2332lim

+

+

x

xx

x

Bi 8. Tnh cc gii hn sau:

1. xxx

+

1lim 2 2. 2lim ( 3 )x

x x x

+

3. )xxxxx

914lim 22 ++

4. 32 2lim ( 3 )x

x x x

+

5. 34412lim 2 ++++

xxxx

6. )13lim 3 23 ++

xxxxx

7. 2lim ( 2)x

x x x

+ + + 8. xxxxx

22lim 23 23 ++


1.0

lim . cotx

x x

2.1

lim(1 ).tan2x

xx

3.3

lim(4 ).sinx

xx

+ 4.0

lims in2 .cot6x

x x


39

Bi 7: Cho tdin SABC. Gi M,N l cc im trn cc onSB v SC sao cho MN khng song song vi BC . Tm giaotuyn ca mt phng (AMN ) v (ABC), mt phng (ABN ) v

(ACM).Bi 8: Cho tdin ABCD. Trn AB ly M vi AM =

3

1AB. Gi

I, K ln lt l trung im ca AC, AD. nh giao tuyn (d) camt phng (MIK) v (BCD).Bi 9: Cho tdin SABC. Gi I, J, K l ba im tu trn SB,AB, BC sao cho JK khng song song vi AC v SA khng songsong vi IJ. nh giao tuyn ca (IJK) v (SAC).Bi 10: Cho t din ABCD. Gi E, F, G l ba im trn AB,AC, BD sao cho (EF) ct (BC) ti I , (EG) ct (AD) ti H. nhgiao tuyn ca mt phng (EFG) vi hai mt phng (BCD) v(ACD).

CHNG MINH BA IM THNG HNG.

Bi 1: Cho tdin SABC. Trn SA, SB v SC ln lt ly ccim D, E, F sao cho DE ct AB ti I, EF ct BC ti J, FD ct

CA ti K. Chng minh rng ba im I, J, K thng hng.

Bi 2: Cho hai mt phng (P) v (Q) ct nhau theo giao tuyn d.

Trong (P) ly hai im A v B sao cho AB ct d ti I. O l mt

im nm ngoi (P) v (Q) sao cho OA v OB ln lt ct (Q)

ti A

v B

.1. Chng minh ba im I, A, Bthng hng.

2. Trong (P) ly im C sao cho A, B, C khng thng hng.

GisOC ct (Q) ti C, BC ct BCti J, CA ct CAti K.

Chng minh I, J, K thng hng.

Bi 8: Cho tdin SABC c D, E ln lt l trung im AC,

BC v G l trng tm tam gic ABC. Mt phng (P) qua AC ct

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38

CHNG II. QUAN HSONG SONG

TM GIAO TUYN CA HAI MT PHNG.

Bi 1: Cho S l mt im khng thuc mt phng hnh bnhhnh ABCD. Tm giao tuyn ca hai mt phng (SAC) v(SBD).Bi 2: Trong mt phng ( ) cho t gic ABCD sao cho AB,

CD khng song song. S l im nm ngoi ( ) . Tm giao tuynca cc cp mt phng sau:

1.

(SAC) v (SBD).2. (SAB) v (SCD).

Bi 3: Trong mt phng ( ) cho ba im A, B, C. S l im

khng thuc ( ) . M, N , I ln lt l trung im ca AB, BC,SA.

1. Tm giao tuyn ca (SAN ) v (SCM).

2.

Tm giao tuyn ca (SCM) v (BIC).Bi 4: Cho t din ABCD v im M thuc min trong caACD . Gi I v J tng ng l hai im trn cnh BC v BD

sao cho IJ khng song song vi CD.1. Hy xc nh giao tuyn ca hai mt phng (IJM) v

(ACD).2. Ly N l im thuc min trong ca ABD sao cho JN

ct on AB ti L. Tm giao tuyn ca hai mt phng (MN J) v

(ABC).Bi 5: Cho hnh chp S.ABCD c y l tgic ABCD c haicnh i din khng song song. Ly im M thuc min trongca SCD . Tm giao tuyn ca hai mt phng:

1. (SBM) v (SCD).2. (ABM) v (SCD).3. (ABM) v (SAC).

Bi 6: Cho tdin ABCD. M v N ln lt l trung im AD v

BC. Tm giao tuyn ca hai mt phng (MBC) v (N AD).


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HM SLIN TC

Bi 1. Xt tnh lin tc cc hm ssau ti 0x :

1. ( )3 11

2 1

+ =

=

xkhi x

f x x

khi x

ti 0 1x =

2. ( )

3 82

25 2

= =

xkhi x

f x x

khi x

ti 0 2x =

3.2

55

( ) 2 1 3

( 5) 5

= =

xkhi x

f x x

x khi x

ti x0= 5

4.

3

2

1 cos0

sin( )

1 02

xkhi x

xf x

khi x

=

=

ti x0= 0

5. ( )

2 22

25 2

>

=

x xkhi x

f x x

x khi x

ti 0 2x =

6. ( )

11

2 12 1


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28

8.

2

2

3 21

1

1( ) 14

11

6 7

+ >

= =


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Bi 2.Cho hnh chnht ABCD. Gi E,F,H,I ln lt l trungim ca AB,CD,BC,EF. Hy tm mt php di hnh bin tamgic AEI thnh tam gic FCH.

PHP VNT

Bi 1.Xc nh nh ca im A(4,-5) qua php vt tm I(-2;6), ts-2.Bi 2. Cho im M(-1;5) v ng thng d: 2x-3y-8=0. Xcnh nh ca M v d qua php vttm O tsbng 2.

Bi 3. Cho im I(2;-1) v im J(7:4). Tm tm v t ca 2ng trn (C)(I;2) v ng trn (C)(J;3).Bi 4.Cho tam gic OMN . Dng nh ca M, N qua php vttm O, tsk trong cc trng hp sau:

1. 3k= 2.1

2k= 3.

3

4k=

Bi 5.Tm php vt bin:

1.

: 12 4x yd = thnh ' : 2 6 0d x y = .

2. 2 2( ) : ( 4) 2C x y+ + = thnh 2 2( ') : ( 2) ( 3) 8C x y + =

PHP N G DN G

Bi 1.Cho im A(3;-4) v ng thng d: 9x+y-6=0 . Vit pt

ng thng d l nh ca d qua php ng dng c c bngcch thc hin lin tip php Ov php V(A,1/3).Bi 2.Cho ng trn (C) c tm I(-1;3), bn knh bng 2. Vitphng trnh ng trn nh ca (C) qua php ng dng cc tvic thc hin lin tip php V(O,3)v php Oy.Bi 3.Cho hnh vung ABCD tm O, M l trung im cnh AB.Xc nh php ng dng bin OAM thnh DBC.

BI TP N CHNG


29

3

12

4( )

3 2 2 22

ax khi x

f x

x khi xx

+

=+

>

Bi 5. Chng minh phng trnh:1. 22 6 1 0 + =x x c t nht 2 nghim.2. 32 10 7 0 =x x c t nht hai nghim.3. cos 0 =x x c nghim.

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Trng THPT N g Thi N him Bi tp ton 11 Trng THPT N g Thi N him Bi tp ton 11


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CHNG IV. O HM

Bi 1.Tnh o hm cc hm ssau:1. 5 3 24= +y x x x x 2.

36= +y x

x

3. 21 1

= + +y xx x

4.3

2 3= + +y x

x x

5. ( )( )3 2 24 2 7= y x x x x 6. 2( 2) 1= +y x x

7.2 1

=+

xy

x 8. 2 3

4=

+

xy

x

9.25 3

2

=

x xy

x 10.

1

1

=

+

xy

x

11. ( )37 25= y x x 12. ( )

52 33 4= + +y x x x

Bi 2. Tnh o hm cc hm ssau:

1.sin

= x

yx

2. cos1

=+

xy

x

3.sin cos

sin cos

+=

x xy

x x 4. sin3 cos tan

5= + +

xy x x

5.

( )

102 sin5= y x x 6. 3sin 3=y x

7. 2cot 1= +y x x 8. ( )2sin cos 2=y x

9. 2 sin 3=y x x 10. 3 2sin 1= +y x

11. 2 2tan 3 cot 2= +y x x 12. sin 4 cos .= +y x x x Bi 3.Gii cc bt phng trnh:

1. ( )' 0>

f x vi ( )

2 5 4

2

x x

f x x

+=


35

PHP QUAY

Bi 1.Cho hnh vung ABCD c tm O. Tm nh ca tam gic

OAB qua php quay tm C gc quay -900

Bi 2. Tm to cc im nh ca A(-3;4), B(-5;1), C(-2;3)qua php quay Q(O,90

o)

Bi 3. Cho im M(3;-4) v ng thng d: 6x-y+10=0. Xcnh nh ca M v d qua php quay tm O mt gc 900.Bi 4. Trong mt phng Oxy, tm php quay Q bin im A(-1,5) thnh im B(5,1).Bi 5. Trong mt phng Oxy, cho im A(0,3). Tm

B = Q (A)(O ; 45 )

Bi 6. Trong mt phng Oxy, cho ng trn2 2(C) : (x 3) (y 2) 4 + = . Tm (C ) = Q (C)

(O ; 90 )

Bi 7.Trong mt phng ta Oxycho ng trn c phng

trnh : 2 2 2 4 4 0x y x y+ + = . Vit phng trnh ng trn

l nh ca ng trn cho qua php quay tm O gc quay090 ; - 090

PHP DI HN H

Bi 1. Cho 2 im A(-2;1), B(3;5) v ng thng d: 4x-9y+6=0.

1. Xc nh nh ca im A qua php di hnh c c bngcch thc hin lin tip php Q(O,90o)v php B.

2. Xc nh nh ca im B qua php di hnh c c bngcch thc hin lin tip php Av Oy.

3. Xc nh nh ca im A qua php di hnh c c bngcch thc hin lin tip php Q(O,90

o)v php d.

4. Xc nh nh ca d qua php di hnh c c bng cchthc hin lin tip php Ov tnh tin

ABT

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Bi 1. Trong mt phng Oxy, hy tm nh ca im M( 2, 1)qua php i xng trc Ox, ri i xng trc Oy.

Bi 2.Trong mt phng Oxy, cho im A(-5; 6), ng thng d:2x-3y-1=0 v ng trn (C): (x-1)2+(y+2)2= 25.1. Xc nh nh ca A v ng thng d qua php Ox.2. Xc nh ng trn (Co) sao cho (C) l nh ca (Co)

qua php Oy.3. Xc nh nh ca (C) qua php d.

Bi 3. Cho im M(2;-7) v ng cong (C) c phng trnhy = x3+3x2-2x+1 .

1.

Xc nh tocc nh ca im M qua php Ox, Oy.2. Vit phng trnh ng cong (C) l nh ca (C) qua

php Ox.Bi 4. Trong mt phng Oxy, cho ng thng( ) : x 5y 7 = 0 + v ( ) : 5x y 13 = 0 . Tm php ixng qua trc bin ( ) thnh ( ) .Bi 5. Cho tgic ABCD. Goi O l giao im ca AC v BD.

Xc nh nh ca tam gic AOB qua php i xng trc CD.PHP I XN G TM

Bi 1.Cho 2 im M(-2;9), N (1;4). Xc nh cc im M1,M2ln lt l nh ca M qua php O , N .Bi 2. Cho im I(-4;3), ng thng d: x-2y+5=0 v ngtrn (C):x2 + y2 -2x+6y+1=0.

1.

Xc nh nh ca I, d v (C) qua php O.2. Vit phng trnh ng thng D l nh ca D quaphp I.

3. Vit phng trnh ng trn (C) l nh ca (C) quaphp I.

Bi 3. Chng minh rng2 2

2 2( ) : 1

x yE

a b+ = v

2 2

2 2( ) : 1

x yH

a b =

c tm i xng l gc ta O.


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2. ( )' 0g x vi ( ) 22 1

1

xg x

x

=

+

3. ' 0y vi

2

2

3

1

xy x

+= +

4. ' 0y vi2

2 1

4

xy

x x

=

+ +

Bi 4.Chng minh rng:1. Hm s xy tan= tha hthc: .01 '2 =+ yy

2. Hm s xy 2cot= tha hthc : .022 '2 =++ yy

3. Hm s 34

xyx

=+

tha hthc : ( ) ( )2

2 ' 1 ''y y y= .

Bi 5.Tnh o hm cp hai ca cc hm ssau:1. 2 cosy x x x= +

2. ( )2 1 tany x x= +

3. 2cos=y x

4. 4 cos2= y x x Bi 6. Vit phng trnh tip tuyn ca thhm s:

1.1

2

=

+

xy

xti im c honh bng 4.

2.2 2 6

1

+ =

x xy

xbit c honh tip im l 3.

3. 3 21 13= + +y x x x bit hsgc k = -3.

4.2 2

2

=

+

x xy

xbit tip tuyn song song vi ng thng

xy 32 = .

5. 3 21

13

= + +y x x x bit tip tuyn vung gc vi ng

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thng: 54

1+= xy .

6. 4 23 4y x x= bit tip tuyn i qua im A(0, -4).

Bi 7. Cho hm s113

+=

x

xy c thl (C). Vit phng trnh

tip tuyn ca (C) sao cho tip tuyn :

1. C tung tip im l 2.2. Vung gc vi ng thng: 4 10y x= + .

Bi 8.Qua im4 4

( , )9 3

A c thkc bao nhiu tip tuyn

n thhm s3

22 33

xy x x= + . Vit phng trnh cc tip

tuyn .


33

PHN II. HNH HC

CHNG I. PHP DI HNH V PHP NG DNGTRONG MT PHNG

PHP TNN H TIN

Bi 1.Trong mt phng Oxy, cho im A(-1,2), B(0,1), C(3,-1)v vect ( 2,3)v=

. Hy tm nh ca cc im trn qua php

tnh tin theo vect v .Bi 2.Trong mt phng Oxy, cho ng thng d: 2x 3y +1 =0 v ng trn (C): x2 + y2- 2x - 4y 4 =0. Hy tm nh ca dv (C) qua php tnh tin theo vect (1, 2)v=

.

Bi 3.Trong mt phng Oxy, cho ng thng ct Ox ti A(-1, 0) v ct Oy ti B(0 ,2). Hy tm nh ca qua php tnhtin theo vect u = (2; 1)

.

Bi 4.Hy tm nh ca ng trn 2 2(C) : (x 3) (y 2) 1 + + = qua php tnh tin theo vect u = ( 2;4)

.

Bi 5. Trong mt phng Oxy, cho A( 5;2) , C( 1;0) . Bit

B = T (A) , C = T (B)u v . Tm u va v

c thbin A thnh C.

Bi 6.Cho ABC c trng tm G. Dng nh ca :1. on thng AB qua php tnh tin

GCT

2. ABC qua php tnh tin 2AGT

Bi 7.Cho hnh bnh hnh ABCD c tm O. Xc nh :1. nh ca ABD qua php tnh tin

3OCT

2. im E sao cho php tnh tinAC

T bin E thnh D.

PHP I XN G TRC

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Bai Tap Toan Lop 11(Full)