Trích Đoạn Cuốn Toán Tập 2

Truy cp website: http://lovebook.vn/ hoc gi ti s in thoi: 0466 860 849 c th s hu cun sch ny.

1 | LOVEBOOK.VN

Trch on cun Tuyn tp 90 thi th km li gii chi tit v bnh lun c i ng tc gi th khoa, gii

quc gia GSTT GROUP bin son do Lovebook.vn sn xut.

Thi gian thm thot thoi a, cun siu phm (ci tn do cc em hc sinh tng) cho i c gn 3 thng.

Trong 3 thng qua, chng ti nhn c rt nhiu nhng phn hi gp t cc em hc sinh v cc thy c khp c

nc:

Theo thy Nguyn Minh Tun - GV chuyn Ha - THPT Hng Vng - Ph Th [tc gi ca hn 20 u sch n thi i

hc ni ting v nhiu ti liu cha s trn mng): y thc s l mt cun sch n thi i hc cht nht, cng phu v

tm huyt nht m thy tng bit ti. Mt hc sinh n thi i hc m khng s hu cun ny th s thit thi rt nhiu

so vi cc bn.

Theo em L Nht Duy [THPT TP Cao Lnh ng Thp]: y l ln u tin em c c mt cun sch tm huyt

nh th ny. Tng li bnh ca anh ch GSTT GROUP rt cht v gn gi na. K t khi cm trn tay cun sch ny,

em cm thy t tin v yu mn ton hn nhiu.

Theo c L Th Bnh [Thc s Ton - Ha] - ging vin khoa Ton Tin ng dng- H Kin Trc H Ni: "Mt cun sch

ng cp v thit thc nht ti tng bit. Khng ch dng li nhng li gii kho khan m cun sch cn cho ta nhng

li t duy, nhng kinh nghim sng mu m h tri qua".

Theo Nguyn Vn Tin [cu hc sinh L Thi T - Bc Ninh, tn sinh vin Y H Ni 29/30]: Lovebook lun bit cch

to ra nhng n phm tht hu ch cho cc em hc sinh, c bit cun Ton. Nm va ri mnh ch tic l cha c cun

Ton, nu c th chc kt qu ca mnh s trn vn hn. Tuy nhin vi 2 cun Ha nm ngoi cng khin mnh t

c c m vo i hc Y H Ni".

Cun tp 2 gm 45 thi i hc c chn lc v tng hp t cc thi th trng chuyn trn c nc trong nm hc 2013 2014. Ngoi ra cun sch cn c khong gn 300 bi ton luyn thm sau mi bi tp in hnh cho cc em luyn.

Khng ch c th cun sch cn bao gm 9 bi phn tch v d on i hc 2014. Vi phn d on ny, cc em c th nm bt c tng quan cc chuyn trong thi chnh thc qua cc nm ca B Gio Dc v c nhng d on tng i chnh xc v dng bi trong thi nm nay, qua vic n tp s trng tm v hiu qu hn.

Cui sch cn c thm 2 chuyn cc cht do i ng tc gi vit na.

nm bt ton b ni dung b TUYN TP 90 THI TH ch trong thng cui, mi cc bn tham gi kha hc c bit ca trung tm VEDU: http://vedu.edu.vn/

NH SCH GIO DC LOVEBOOK

Web: lovebook.vn

Facebook: https://www.facebook.com/Lovebook.vn?bookmark_t=page

Gmail: [email protected]

ST: 0466.860.849. a ch: 101 Nguyn Ngc Ni, Thanh Xun, H Ni

Tuyn tp 90 thi th i hc km li gii chi tit v bnh lun mn Ton tp 2- LOVEBOOK.VN

LOVEBOOK.VN | 2

Phn I: THI + LI GII CHI TIT V BNH LUN S 01

I. PHN CHUNG CHO TT C CC TH SINH (7,0 im)

Cu 1 (2,0 im). Cho hm s 2x 1

yx 1

(1), c th (C).

1. Kho st s bin thin v v th (C) ca hm s (1).

2. Mt hnh ch nht MNPQ c cnh PQ nm trn ng thng : 3x y 11 = 0, hai im M, N thuc (C) v

di ng cho ca hnh ch nht bng 5 2 . Lp phng trnh ng thng MN.

Cu 2 (1,0 im). Gii phng trnh 2sin xsin2x 11cos x cot x

2cot x 3sin2x

(x ).

Cu 3 (1,0 im). Gii phng trnh 1 1

x x ln x 14x 4x

(x ).

Cu 4 (1,0 im). Tnh tch phn I =

x5

x2

e 3x 2 x 1dx

e x 1 x 1

.

Cu 5 (1,0 im). Cho khi t din ABCD c AC = AD = 3 2 , BC = BD = 3, khong cch t nh B n mt phng

(ACD) bng 3 , th tch ca khi t din ABCD l 15 . Tnh gc gia hai mt phng (ACD) v (BCD).

Cu 6 (1,0 im). Tm m phng trnh sau c 3 nghim thc phn bit:

3 x 1 1 x 3 x 1 x 3 m 3 x 1 .

II. PHN RING (3,0 im). Th sinh ch c lm mt trong hai phn (phn A hoc phn B)

A. Theo Chng trnh Chun

Cu 7.a (1,0 im). Trong mt phng vi h trc ta Oxy, cho ng thng d: x + y 2 = 0 v im M(3; 0).

ng thng qua M ct ng thng d ti A. Gi H l hnh chiu vung gc ca A ln Ox. Vit phng trnh

ng thng , bit khong cch t H n bng 2

5.

Cu 8.a (1,0 im). Trong khng gian vi h trc ta Oxyz, cho cc im A(2; 0; 0), B(0; 4; 0), C(0; 0; 3) v D(1;

2; 3). Vit phng trnh mt phng (P) cha AD sao cho tng khong cch t B v C n (P) l ln nht.

Cu 9.a (1,0 im). Gi z1, z2 ln lt l hai nghim ca phng trnh 2z 1 3i z 2 2i 0 v tha mn 1 2z z

. Tm gi tr ca biu thc 2 21 1

1 2A z 1 z

.

B. Theo chng trnh Nng cao

Cu 7.b (1,0 im). Trong mt phng vi h trc ta Oxy, cho hnh thang OABC (OA // BC) c din tch bng 6,

nh A(1; 2), nh B thuc ng thng d1: x + y + 1 = 0 v nh C thuc ng thng d2: 3x + y + 2 = 0. Tm

ta cc nh B, C.

Cu 8.b (1,0 im). Trong khng gian vi h trc ta Oxyz, cho tam gic ABC c C(3; 2; 3), ng cao qua A v

ng phn gic trong gc B ca tam gic ABC ln lt c phng trnh l 1x 2 y 3 z 3

d :1 1 2

v

2

x 1 y 4 z 3d :

1 2 1

. Lp phng trnh ng thng BC v tnh din tch ca tam gic ABC.

Cu 9.b (1,0 im). Gii h phng trnh 2 2

2z w zw 7

z w 2w 2

z,w .


3 | LOVEBOOK.VN

LI GII CHI TIT V BNH LUN

Cu 1.

1.

Tp xc nh: = \ {1}.

S bin thin:

S bin thin:

2

3y 0

x 1

vi mi x .

Hm s nghch bin trn cc khong (; 1) v (1; +).

Gii hn, tim cn:x xlim y lim y 2

; x 1lim y

;

x 1lim y

.

th hm s nhn ng thng x = 1 lm tim cn ng v nhn ng thng y = 2 lm tim cn ngang.

Bng bin thin:

th:

th (C) ca hm s ct trc tung ti im (0; 1), ct trc honh

ti im 1

;02

. ng thi (C) nhn giao im ca hai ng

tim cn l I(1; 2) l trc i xng.

2.

nh hng: u tin vi d kin MNPQ l hnh ch nht th ta khai

thc ngay c tnh cht song song, l MN // PQ. Lc ny ta s

c ngay dng ca phng trnh ng thng MN l

3x y + m = 0, vi m 11, tng ng vi MN: y = 3x + m. Nh

vy honh M v N chnh l nghim ca phng trnh giao im

ca ng thng vi th (C) dng c phng trnh

honh v dng nh l Vit biu din c tng v tch xM

+

xN; x

Mx

N theo bin m.

Tip theo, vi hai ng thng song song th ta lun xc nh c khong cch gia hai ng thng , bi

khong cch gia hai ng thng song song chnh bng khong cch ca mt im bt k trn ng thng ny

n ng thng kia. Trn th ta lun ly c mt im K c ta xc nh dng khong cch s tnh c

khong cch t K n MN di cnh PN = d(K, MN) (theo mt n m).

Vy d kin cui cng l d kin ng cho. V ta c tng v tch xM

+ xN, x

Mx

N theo bin m nn vic tnh di

MN theo m l iu d dng. Ngoi ra, dng nh l Pytago ta s c ngay: MN2 + NP2 = PM2 = 2

5 2 , t y gii

phng trnh n m duy nht tm m MN.

Theo nh hng kh r rng trn ta c li gii:

Bi gii:

Do MNPQ l hnh ch nht nn MN // PQ ng thng MN c dng 3x y + m = 0 y = 3x + m.

x O

1

2

y

I

M

N

P

Q

5 2

K

x + 1

y

+

2 2

y


LOVEBOOK.VN | 4

Phng trnh honh giao im ca ng thng MN v (C) l:

2x 1

3x m 2x 1 x 1 3x mx 1

(d thy x = 1 khng tha mn) 23x m 5 x m 1 0 (*).

(*) c bit thc = 2 2m 5 4.3 m 1 m 2m 37 0 vi mi x (*) lun c hai nghim phn bit

x1, x

2. Theo nh l Vit:

1 2

1 2

5 mx x

3

m 1x x

3

Khng mt tnh tng qut, gi s M(x1; 3x

1 + m) v N(x

2; 3x

2 + m) th

MN2 = 10(x

1 x

2)

2 = 10

2

1 2 1 2x x 4x x

= 10

25 m m 1

4.3 3

= 210 m 2m 379

.

K(0; 11) d(K, MN) =

22

3.0 11 m

3 1

= m 11

10

NP2 = d2(K, MN) =

2

m 11

10

.

p dng nh l Pytago, ta c: MN2 + NP2 = PM2

2

22

m 1m 1110

m 2m 37 5 2 2899 10 m

109

i chiu iu kin m 11, ta c hai gi tr cn tm ca m l m = 1 v m = 289

109

.

Cu 2.

nh hng: Khi nh gi qua phng trnh ny th ta thy rng n cng khng phc tp qu, ch cha hm sin,

cos v cot dng thun (n gin). Nhm trong u nhn t th thy cotx = cos x

sin x; sin2x = 2sinxcosx th thy

ngay c t v mu u xut hin nhn t l cosx.

Tip tc nhp thm t sau khi rt gn cosx t v mu th c:

12sin x.2sin x 11

sin x 21

3.2sinxsin x

, vng, v n y

th phng trnh cng l bn cht ca n: y thc cht l phng trnh mt n t = sinx.

Bi gii:

iu kin:

00

xx

x

2

0 1x

x1

x 6 x 0x 3 0 6x

x 0

sincos sincot

sincos

sinsin

sin

sin (*).

Phng trnh cho tng ng vi:

cos x 12sin x.2sin xcos x 11cos x 2sin x.2sin x 11

sin x sin x2 2cos x 1

3.2sinxcos x 3.2sinxsin x sin x

(do cosx 0).

2 3 21 14sin x 11 2 6sin x 4sin x 12sin x 11sin x 3 0sin x sin x


5 | LOVEBOOK.VN

2sin x 1 2sin x 3 sin x 1 0

x k2

x k2

6

5

21

x2

x

x k2

1

6

sin

sin

(k ).

Th li (*), ta c phng trnh c hai h nghim l x =

6 + k2 v x =

5

6 + k2 (k ).

Cu 3.

nh hng: u tin, iu kin x > 0 l khng th thiu.

Nhn thy phng trnh c cha hm hu t v c hm logarit (hai hm khc tnh cht) nn ta ngh ngay n

phng php hm s trong u.

nh hng u tin gip ta pht trin hng gii cho bi ton: Chng ta nn dng hm s theo kiu tnh n iu

hay l nn dng hm s theo kiu hm g(f(x)) = g(h(x)), vi g l hm n iu?

Nu trin khai theo hng th nht: vic o hm trnh phc tp, chng ta s nn chia hai v cho x. Bi v ta

ly o hm ca 1

x.ln x4x

th s phc tp hn so vi vic ly o hm ca ln

1x

4x

.

Nh vy chia hai v cho x ta c: 2 2

1 1 1 1 1 11 ln x 1 ln x 0

4x x x 4x4x 4x

(*).

Th ly o hm ca v tri ta c:

3 2

3

2

3 2 2

11

1 1 4x1

2x 1 4x

2x x

6x 1

2x 1 4x4x

x

.

Vy vic dng hm n iu ca chng ta tiu tan khi m o hm khng dng hoc khng m vi

x > 0. Nhng nh th cng ng vi nn nh , khi o hm c nghim (v ch c mt nghim p) th ta c th

v c bng bin thin ca hm s, v bit u n s c nghim p cho chng ta nhn xt!

Tht vy, th lp bng bin thin th thy ngay VT(*) 0. Du ng thc xy ra khi x = 1

2 (chnh l nghim ca

o hm lun!).

Nu trin khai theo hng dng hm s. Cch ny s thng c cc bn ngi (ni ng hn l li) o

hm dng!

Khi gp phng trnh dng: A(x)

A(x) ln B(x)B(x)

(vi A, B dng) th ta bin i mt cht phng trnh s

thnh: A(x) ln A(x) B(x) ln B(x) , phng trnh ny c dng hm ng bin l f(t) = t + lnt, l hm ng

bin trn (0; +).

Vy khi gp phng trnh ny th ta thy trong logarit c th phn tch c thnh nhn t, ng thi mun a

phng trnh v c dng trn th u tin mnh phi chia hai v cho x . Ta thu c phng trnh:

2

2 2 2 2

11

1 1 1 1 1 1 1 1 14x1 ln x 0 1 ln 1 ln 1 ln

x 4x 1 x x x4x 4x 4x 4xx

.

n y th dng hm s xut hin v vic cn li ca chng ta cng khng qu kh na!

Bi gii:

Cch 1.

iu kin x > 0. Phng trnh cho tng ng vi:


LOVEBOOK.VN | 6

2 2 2

1 1 1 1 1 11 ln x 1 ln x 1

4x x x4x 4x 4x

2 2 2 2

1 1 1 1 1 1 11 ln 1 ln x 1 ln 1 ln

x x x4x 4x 4x 4x

(*).

Xt hm s f(t) = t + lnt trn (0; +). Ta c: 1

f (t) 1 0t

vi mi t > 0 f(t) ng bin trn (0; +).

Mt khc (*) c dng 2

1 1f 1 f

x4x

(vi

2

11 0

4x v

10

x )

2

2

1 1 1 11 1 0 x

x 2x 24x

.

Vy nghim ca phng trnh l x = 1

2.

Cch 2.

iu kin x > 0. Chia hai v ca phng trnh cho x ta c:

2 2

1 1 1 1 1 11 ln x 1 ln x 0

4x x x 4x4x 4x

.

Xt hm s f(x) = 2

1 1 11 ln x

x 4x4x

trn (0; +).

Ta c:

2

3

3 2

3 22

11

1 1 4x1

2x 1 4x

2x

6x 1f (x)

x x4x

2x 1 4x

;

1f (x) 0 x

2 (do x > 0).

Lp bng bin thin cho ta f(x) 0 vi mi x > 0. Ta c f(x) = 0 x = 1

2.

Vy nghim ca phng trnh l x = 1

2.

Bi tp cng c:

Gii phng trnh: x x1969

2014 x ln 19692014

(p s: x = 0).

Cu 4.

nh hng: Nhn thy tch phn c cha c hm v t, hu t v c hm m (cc hm khc tnh cht) nn ta ngh

n phng php tch phn tng phn, hoc tc dng I = b b

a a

g(x)f(x)

g(x) lm d dng hn. Nhng vi bi ton

th cch dng tch phn tng phn gn nh v hiu. Vy nn ta suy ngh n hng th hai l tch I thnh dng

nh trn. Mt iu gi cho chng ta thc hin theo phng n th hai na l t s c phn ging vi mu s

(phi ni l rt ging), nn vic rt gn bt i l iu ng nhin:

x x

x x

e 3x 2 x 1 e 2x 11

e x 1 x 1 e x 1 x 1

.

Nh vy s 1 tch ra th d dng ly nguyn hm, cn lng

x

x

e 2x 1

e x 1 x 1

th vn cha c dng

g(x)

g(x). Vy

phi lm sao? Khng l li b cuc gia chng? ng lo, khi cha gp dng ny th mun xut hin dng g(x)

g(x) th

nhiu lc ta phi cng chia c t c mu cho mt lng no (v thng th lng ny l lng tng ng,


7 | LOVEBOOK.VN

hoc l nhn t mu s hoc t s), hoc c lc l nhn c t v mu vi mt lng no xut hin c

dng . Th xem nh!

Vi cc din nh th ny th ta s c hai hng:

+ Hng 1: Chia hai v cho ex ta c:

x

2x 1

x 1x 1

e

cng cha thy xut hin dng g(x)

g(x).

+ Hng 2: Chia hai v cho x 1 ta c:

x

x

e 2x 1

x 1

e x 1 1

. Th ly o hm mu

xx e 2x 1e x 1x 1

(chnh

bng t s), thnh cng!

Bi gii:

Ta c:

x5 5

x2 2

e 2x 1I dx dx

e x 1 x 1

.

+) 5

5

1 22

I dx x 5 2 3 .

+)

x

x5 5 55

x2 2x x 2

2 2

e 2x 1

e x 1 1 2e 12 x 1I 2 dx 2 dx 2ln e x 1 1 2lne 1e x 1 1 e x 1 1

'

.

Vy 5

1 2 2

2e 1I I I 3 ln

e 1

.

Cu 5.

nh hng: T din ABCD ta bit c di 4 cnh, v li c iu c bit l A v B u cch u hai im C,

D (AC = AD, BC = BD) A, B nm trn mt phng trung trc ca cnh CD. V mt phng trung trc ny chnh l

mt phng i qua A, B v trung im M ca CD gc gia hai mt phng (ACD) v (BCD) chnh bng AMB hoc

bng 0180 AMB (ty vo ln gc AMB l nh hn 900 hay ln hn 900). ng thi bi ra cn cho thm khong cch gia mt nh n mt phng i din v cho thm c th tch khi t

din d dng tnh c din tch mt y l ACD tnh c di CD (do ACD bit di 2 cnh)

BCD hon ton xc nh cc thng s v 3 cnh tnh c BM (l ng cao BCD).

Ngoi ra nhn thy c khong cch t B n (ACD) nn sin

(ACD) (BCD), = d B (ACD)

BM

, t xc nh c gc

gia hai mt phng (ACD) (BCD), .

Bi gii:

Theo bi ra: d(B, (ACD)) = 3 ; VABCD

= 15 (vtt).

Ta c: SACD

=

ABCD3V

d B (ACD), =

3 15

3 = 3 5 (vdt).

Mt khc: SACD

= 1

2AC.AD.sin CAD

H

A

B

C

D

M


LOVEBOOK.VN | 8

sin CAD = DAC2S

AC.AD =

2.3 5

3 2.3 2 =

5

3.

cos CAD = 21 sin CAD = 2

3.

Gi M l trung im ca CD th do ACD cn ti A v BCD cn

ti B nn BM CD v AM CD (ABM) (ACD). Gi H l

hnh chiu ca B ln (ACD) th ta c H thuc ng thng AM,

ng thi di BH = d(B, (ACD)) = 3 . Ta c gc gia mt

phng (BCD) v (ACD) chnh bng BMH < 900.

+) Trng hp 1:

cos CAD = 2

3 CD = 2 2AC AD 2AC.ADcosCAD = 2 3

BM =

22

2 2CD 2 3BC 3 62 2

.

sin BMH = BH

BM =

3

6 BMH = 450.

+) Trng hp 2:

cos CAD = 2

3

. Tng t ta tnh c CD = 2 15 > BC + BD, khng tha mn bt ng thc tam gic loi.

Vy gc gia hai mt phng (BCD) v (ACD) l 450.

Lu : C th xy ra hai trng hp v v tr im H nh 2 hnh v trn, nhng d th no i na th gc gia hai

mt phng (BCD) v (ACD) vn bng 450.

Cu 6:

nh hng: tng v nhng bi tm m phng trnh c nghim l khng xa l g na tng ca chng

ta l c lp m thu c dng m = f(x), sau kho st f(x) kt lun cc gi tr ca m tha mn iu kin

bi.

Vi bi ny, mun c lp m mt cch nhanh chng th ta chia hai v cho 3 x 1 . Th nhng trc khi chia th

ta phi xt trng hp x = 2 ( m bo 3 x 1 0). Khi ta th x = 2 vo v tri th thy rng v tri cng bng 0 chc chn v tri c th phn tch c nhn t (x 2) nhn t (x 2) c th chia c cho

3 x 1 (v c hai u c nghim bng x = 2). Tht vy:

x 2 = 3 x 1 3 x 1 3 x 1 .

Vy nn ta chn cch thun li hn cho li gii l phn tch v tri cha nhn t 3 x 1 bi gii c ngn gn hn!

VT = 3 x 3 x 1 1 x x 2 3 x 3 x 1 1 x 1 3 x

3 x 1 3 x 1 x 1 3 x .

Nh vy chuyn v ta s thu c hai nhn t l 3 x 1 v 3 x 1 x 1 x 3 x m 3 .

H

A

B

C

D M


9 | LOVEBOOK.VN

Ci kh cn li l i x l nhn t th hai:

3 x 1 x 1 x 3 x m 3 0 m 1 x 3 x 1 x 3 x 3 (1).

X l phng trnh ny cng khng h kh, thng th ta s t 2t 1 x 3 x t 4 2 1 x 3 x

(1) gn nh c x l. Th nhng vi cc bn thun thc vic gii phng trnh ri th s chn cch

kho st v phi ca (1) lun khng mt thi gian bin lun theo n t na.

Bi gii:

iu kin x1 3 .


3 x 3 x 1 1 x 1 3 x 3 m 3 x 1

3 x 1 3 x 1 x 1 3 x 3 m 3 x 1

3 x 1 3 x 1 x 1 x 3 x m 3 0

x 2

m 1 x 3 x 1 x 3 x 3

(*)

Phng trnh cho c ba nghim phn bit khi v ch khi (*) c hai nghim phn bit khc 2.

Xt hm s f x 1 x 3 x 1 x 3 x 3 trn 1;3 .

Vi mi x 1;3 :

1 1 2x 2f (x) 0

2 1 x 2 3 x 2 1 x 3 x

.

f x 0 1 x 3 x 2x 2

1 x 3 x 1 x 3 x 1 x 3 x

2 7

1 x 3 x 1 x2

.

Bng bin thin:

7

Da vo bng bin thin, kt hp vi iu kin x 2 (v f(2) 2 2 3 ) ta c th kt lun c cc gi tr ca m

cn tm l 11

5;2

m

2 2 3 .

Cu 7.a.

nh hng: i qua im M nn c th vit c phng trnh ng thng dng tng qut:

a(x xM

) + b(y yM

) = 0.

f (x)

3 1 + 2 + 7

2

f(x)

11

2

5 1

x

+ 0


LOVEBOOK.VN | 10

dng sn hai n a, b. Tip tc tm ta im A theo hai n a, b sau chiu A ln Ox c ta im

H v bc cui cng l dng d kin khong cch tm t s a

b phng trnh .

Bi gii:

+) i qua im M(3; 0) nn c phng trnh l:

a(x 3) + b(y 0) = 0 : ax + by 3a = 0 (iu kin a2 + b2 0).

+) Ta A l nghim ca h:

3a 2bxy 2 x y 2 xx y 2 0 a b

ax b 2 x 3a 0 a b x 3a 2bax by 3a 0 3a 2by 2

a b

(iu kin a b).

+) Hnh chiu H ca A ln Ox s c ta l H(xA; 0) H

3a 2b0

a b

; .

+) d(H, ) = 2

22 2 2 2 2 2

2 2

3a 2ba. b.0 3a

a b 2 ab 4a b 5a b 4 a b a b

a b 55a b

2 22 2

a 2b

a 2b 2a b 2a ab 2b 0 b 2a

2a ab 2b 0

Nu a = 2b chn b = 1 a = 2 (tha mn) : 2x + y 6 = 0.

Nu b = 2a chn a = 1 b = 2 (tha mn) : x + 2y 3 = 0.

Vi a b th 22 2 2 21 32a ab 2b a b a b 0

2 2 .

Vy c hai phng trnh ng thng tha mn l 1: 2x + y 6 = 0 v

2: x + 2y 3 = 0.

Cu 8.a.

nh hng: Mt phng (P) i qua hai im c nh A, D bit ta nn c th dng phng php chm mt

phng mt cch gin tip, bng cch gi phng trnh mt phng (P) dng tng qut (s ch c hai n). Vic x

l tng khong cch cc i ta s dng bt ng thc n gin nh Cauchy hay Bunhiacpxki (p dng vi

cc bn kh, gii), hoc cc bn khng quen dng cc bt ng thc th c th dng xt hm s.

Bi gii:

+) Gi s phng trnh (P) l: ax + by + cz + d = 0 (iu kin 2 2 2a b c 0 ).

im A(2; 0; 0) (P) 2a + d = 0 d = 2a.

im D(1; 2; 3) (P) a 2b + 3c + d = 0 c = 2b a d

3

=

2b a

3

.

(P): ax + by + a 2b

3

z 2a = 0.

(hng x l trn chnh l hng x l theo phng php chm mt phng mt cch gin tip).

+) Tng khong cch t B v C n mt phng (P) l:

h = d(B, (P)) + d(C, (P)) = 2 2 2

2 2 2 2 2 2

a 2b3. 2a

4b 2a 3 2b a3

a 2b a 2b a 2ba b a b a b

2 2 2

.

p dng bt ng thc Bunhiacpxki ta c:


11 | LOVEBOOK.VN

h =

2

2 2 2 2 2

2

9. 2b a 9. a 2b 9 a 2b6

3a 2ba 2b

2a 2b 22

a 2b2 2 1 . a b

2

.

ng thc xy ra

a 2ba b 2 a b2 2 1

chn a = 3 b = 3 (P): 3x 3y z 6 = 0.

Nhn xt: Mu cht ca bi ton ny l dng bt ng thc nh th no cho hp l. lm c iu th ta

ch ch n vic dng bt ng thc cho mu s cho hp l.

K thut chn im ri s c cp y .

Ta dng bt ng thc: 2

2 2 2 2 2 a 2 a 1xa ybb

x y 1 a b b x a y 1 b2 2 2

, trong x, y

l cc hng s chng ta tm s dng bt ng thc cho hp l.

ng thc xy ra

a 2ba b a 2b2 x 2y 2x y 1 x 2y

(1).

Mun 1

x a y 1 b2

rt gn c cho t s th ta phi c

1y 1 2 x

2

(2).

Gii h (1), (2) ta c x = 2; y = 2. n y th vic p bt ng thc Bunhiacpxki l iu d dng .

Lu : Vi bi ton ny th cch s dng i s l ti u nht. Vic s dng phng php hnh hc s rt phc tp

trong vic bin lun, dn n ng nhn kt qu bi lm sai.

Cu 9.a.

y l mt bi ton hon ton c bn, ch yu cu bn nm c cch gii phng trnh bc 2 trong tp s phc

l c. Nhng li khuyn cho cc bn l khi tm c nghim ca phng trnh ri th chng di g li trnh by

theo cc bc gii phng trnh mnh lm trong nhp vo giy thi c! Hy dng cch phn tch nhn t trong

bi lm, ta ch cn dng cc du tng ng ch khng cn vit cu ch g nhiu nh .

Bi gii:


2z 2i

z 2i 1 i z 2i 1 i 0 z 2i z i 1 0z i 1

.

Do 1 2

z z nn ta c z1 = 2i v z

2 = i + 1.

Ta c: 2 2 22 21 1 21 1 i 1 3

A 2i 1 i 1 i 12i i 2 2 2

.

Cu 7.b.

nh hng: Do ta ca A v O bit nn phng trnh ng thng OA l hon ton xc nh dng ca

phng trnh ng thng BC (ch cha mt n cn tm l m). Vy hon ton c th xc nh c ta im B

v C theo mt n m, da vo h phng trnh giao im ca ng thng BC vi ng thng d1 (tm c B); h

phng trnh giao im ca ng thng BC vi ng thng d2 (xc nh c C).

Cui cng ta khai thc d kin din tch: S = 1

OA BC d O BC2

. , y s l phng trnh c mt n duy nht

l m tm m ta B, C.

Bi gii:


LOVEBOOK.VN | 12

+) Phng trnh OA: x 0 y 0

2x y 0.1 0 2 0

OA // BC phng trnh ng thng BC c dng: 2x + y + m = 0 (vi m 0).

+) Ta B l nghim ca h: x y 1 0 x 1 m

2x y m 0 y m 2

B(1 m; m 2).

+) Ta C l nghim ca h: 3x y 2 0 x m 2

2x y m 0 y 4 3m

C(m 2; 4 3m).

+) Din tch hnh thang OABC l: S = 1

2(OA + BC).d(O, BC)

2 2 2 22 2

m1( 1) 2 (2m 3) (4m 6) . 6

2 2 1

2m 3 1 m 12 (*).

Phng n ti u nht gii phng trnh ny s l ph du gi tr tuyt i!

Nu m < 0 th (*) thnh: (3 2m + 1).(m) = 12 m2 2m 6 = 0 m = 1 7 .

Kim tra iu kin ta ch ly nghim m = 1 7 B 7 1 7 ; v C 1 7 1 3 7 ; .

Nu 0 < m < 3

2 th (*) thnh: (3 2m + 1).m = 12 m2 2m + 6 = 0, v nghim.

Nu m 3

2 th (*) thnh: (2m 3 + 1).m = 12 m2 m 6 = 0 m = 3 hoc m = 2.

Kim tra iu kin ta ch ly nghim m = 3 B(2; 1) v C(1; 5).

Vy c hai cp im B, C tha mn bi nh trn.

Cu 8.b.

Ta x l bi ton ny ging nh x l mt bi ton hnh hc phng, v phng php th khng c g mi khi gp

ng cao (tn dng yu t vung gc) v ng phn gic (tn dng phng php ly i xng).

Bi gii:

+) d1, d

2 c vct ch phng ln lt l

1u = (1; 1; 2) v

2u = (1; 2; 1).

+) B d2:

x 1 t

y 4 2t

z 3 t

B(1 + t; 4 2t; 3 + t) CB = (t 2; 2 2t; t).

d1 l ng cao k t A nn

1u .CB 0 (t 2) + (2 2t) + (2).t = 0 t = 0 B(1; 4; 3).

BC i qua C v nhn vct 3

1u BC

2 = (1; 1; 0) lm vct ch phng

phng trnh ng thng BC l

x 3 t

y 2 t

z 3

(t ).

+) Gi H(a; b; c) th trung im ca CH thuc d2, ng thi

2CH u nn ta H l nghim ca h:

a 3 b 2 c 3a 11 4 3

2 2 2b 2

1 2 1c 51 a 3 2 b 2 1 c 3 0

. . .

H(1; 2; 5).

+) Thy rng H d2 A H A(1; 2; 5) v ABC vung ti A.


13 | LOVEBOOK.VN

Din tch tam gic ABC l: S = 1

2AB.AC =

1.2 2.2 2

2 = 4 (vdt).

Nhn xt: Bi ton ny s l vt tm thi i hc nu im H tm c khng thuc ng thng d2. Bi nu vy

th sau khi tm c im H, ta s phi i vit phng trnh AB, ri tm ta A dng cng thc din tch

tnh din tch tam gic th bi lm tr nn qu di, khng ph hp vi mt bi thi i hc (nht l cu n im

nh ta khng gian). Vy nn trong qu trnh lm bi, cc bn hy ch n s c bit ca bi, ch ng

di g m c i theo li mn phng php m ta s dng lu nay trong khi gii ton.

Nu gp mt bi tng t th ny th khi tm c ta H, nu thy H d2 th khi dng cng thc tnh din tch,

ta dng S = 1

2AB.CH nh! ng nn dng cng thc S =

1

2BC.d(A, BC) trong trng hp ny v lm nh vy s

phc tp tnh ton hn ch dng cng thc tnh khong cch t mt im n mt ng thng cho trc!

Cu 9.b.

Phng trnh th nht ca h tng ng: w 7

z 2 w w 7 z2 w

(d thy w = 2 khng tha mn).

Th vo phng trnh th hai ca h ta c:

2

2 4 3 2 2 2w 7 w 2w 2 w 6w 15w 2w 57 0 w 7w 19 w w+3 02 w

2

2

2 2

7 3i 3 5 3i 3w z

2 2

7 27 7 3i 3 7 3i 3 5 3i 3w w w zw 7w 19 0 2 4 2 2 2 2

w w 3 0 1 11 1 i 11 3 i 111 11 w w zw2 2 2 22 4

1 i 11 3 i 11w z

2 2

Vy h phng trnh cho c 4 nghim:

(z; w) = 5 3i 3 7 3i 3 5 3i 3 7 3i 3 3 i 11 1 i 11 3 i 11 1 i 11

2 2 2 2 2 2 2 2

; , ; , ; , ; .

Nhn xt: Vic bin i phng trnh bc 4 c nghim thc th khng qu kh khn, c th dng my tnh nhm

nghim v on nhn t chung. Th nhng vi phng trnh bc 4 nghim phc (v khng c nghim thc) th

vic dng my tnh nhm nghim ri on nhn t l khng th. Vy nn ta phi dng k thut gii phng

trnh bc 4 phn tch nhn t chung mt cch nhanh chng:

2

4 3 2 2 2w 6w 15w 2w 57 0 w 3w 6w 2w 57 .

By gi ta thm vo hai v mt lng l 2 22m. w 3w m ( v tri c mt bnh phng ng):

2

2 2 2w 3w+m 2m 6 w 2 1 3m w m 57 (*).

Mun v phi l mt bnh phng ng (hoc c th l lng m ca bnh phng ng: A2) th:

= 0 2 2 77 3 331 3m 2m 6 m 57 0 m 11 m4

.

V l do thm m nn chng ta chn m = 11. Thay m = 11 vo (*):

2 22 2 2 2w 3w+11 16w 64w 64 4w 8 w 7w 19 w w+3 0 .


LOVEBOOK.VN | 14

Bc ny ta ch cn lm ngoi nhp ri rinh vo bi lm nh .


15 | LOVEBOOK.VN

S 2


Cu 1 (2,0 im). Cho hm s y = x 3

x 2

c th (C).

1. Kho st s bin thin v v th (C) ca hm s.

2. Tm cc gi tr thc ca m ng thng (d): y = x + m ct (C) ti hai im phn bit A, B nm hai pha

trc tung sao cho gc AOB nhn (O l gc ta ).

Cu 2 (1,0 im). Gii phng trnh cos2x + sin2x cosx (1 sinx)tanx = 0 (x ).

Cu 3 (1,0 im). Gii bt phng trnh

2

23

x 4x 9x 6

x 4x 3x 1 1

1

2

(x ).


2

3

sin2x cos x 1 2x cos x 1 ln xdx

sin x x ln x

.

Cu 5 (1,0 im). Cho hnh lng tr ng ABC.ABC c y ABC l tam gic cn ti C, cnh AB = 2a v

gc ABC = 300. Mt phng (CAB) to vi mt y (ABC) mt gc 600. Tnh th tch ca khi lng tr ABC.ABC

v tnh khong cch gia hai ng thng AB v CB theo a.

Cu 6 (1,0 im). Cho cc s thc a, b, c thuc on [0; 1]. Chng minh rng:

a b c

1 a 1 b 1 cb c 1

1c a 1 a b 1

.



Cu 7.a (1,0 im). Trong mt phng vi h trc ta Oxy, cho hnh ch nht ABCD c nh A nm trn ng

thng : x y + 1 = 0. ng cho BD c phng trnh: 5x y 7 = 0. Xc nh ta cc nh hnh ch nht

cho, bit rng I(1; 4) l trung im ca CD v nh D c honh l mt s nguyn.

Cu 8.a (1,0 im). Trong khng gian vi h trc ta Oxyz, cho mt cu (S): 2 2 2x y z 2x 4y 4z 16 v

ng thng : x y z 5

1 1 4

. Vit phng trnh (P) cha ng thng v ct mt cu (S) theo mt ng trn

c bn knh bng 4.

Cu 9.a (1,0 im). Anh Thy v ch Hin cng chi Boom Online. V mun tng thm sc hp dn cho tr chi

cng nh s c gng ca mnh, ch Hin ngh ra mt tr c cc: nu ai thng trc 3 vn th thng trn v

ngi thua phi np cho ngi thng 3K. Bit rng s trn chi ti a l 5 vn, xc sut m ch Hin thng mi

trn l 0,4 v khng c trn ha. ng thi khi c ngi thng ng 3 vn ri th tr c cc dng li. Tnh xc

sut m ch Hin s ly c 3K t v thng cc ny?


Cu 7.b (1,0 im). Trong mt phng vi h trc ta Oxy, cho hnh vung ABCD. Gi M l trung im ca cnh

BC, N l im trn on CD sao cho CN = 2DN. Bit ng thng AN c phng trnh: 2x y 3 = 0 v im M c

ta M11

22

; . Tm ta im A.

Cu 8.b (1,0 im). Trong khng gian vi h trc ta Oxyz, cho bn im A(1; 2; 3), B(2; 3; 1), C(0; 1; 1) v

D(4; 3; 5). Lp phng trnh mt phng (P) bit (P) i qua hai im A, B, ng thi C v D cch u (P).

Cu 9.b (1,0 im). Tnh mun ca s phc z, bit rng 3z 12i z v z c phn thc dng.

HT


LOVEBOOK.VN | 16


Cu 1.

1.

Tp xc nh: = \ {2}.

S bin thin:

Chiu bin thin:

2

5y 0

x 2

vi mi x .

Hm s nghch bin trn cc khong (; 2) v (2; +).

Gii hn v tim cn: x xlim y lim y 1

; x 2lim y

; x 2lim y

= .

th hm s nhn ng thng y = 1 lm tim cn ngang v nhn ng thng x = 2 lm tim cn ng.

Bng bin thin:

th:

th (C) ca hm s ct trc tung ti 3

02

; , ct trc

honh ti im (3; 0). ng thi (C) nhn giao im ca

hai ng tim cn I(2; 1) lm tm i xng.

2.

nh hng: Chc chn l trong qu trnh x l bi ton th phi dng n phng trnh honh giao im ca

(C) vi (d). Thy phng trnh honh giao im c dng bc 2 nn vic dng nh l Vit l iu ng nhin!

Gi hai nghim ca phng trnh l x1, x

2 th theo bi ra, x

1 v x

2 phi tri du ac < 0.

Tip tc x l gc AOB nhn. rng AOB chnh l gc hp bi hai vct OA v OB , ng thi thy rng trong

qu trnh gii th ta cha s dng nh l Vit, vy nn ta cn ngh ra mt lin h i xng A, B p dng c

nh l Vit. R rng, AOB nhn cos AOB > 0 (1). Thm mt cht gia v vo hai v: nhn c hai v vi OA.OB

th (1) OA.OB > 0, y chnh l mt lin h i xng vi A, B gip ta s dng c nh l Vit!

Bi gii:

+) Phng trnh honh giao im ca (C) v (d):

x 3

x m x 2 x m x 3x 2

(d thy x = 2 khng l nghim)

2x m 1 x 2m 3 0 (*).

+) d ct (C) ti hai im phn bit A, B nm hai pha trc tung

(*) c hai nghim phn bit x1, x

2 tha mn x

1x

2 < 0

P = 2m + 3 < 0 m < 3

2

(**).

Lc ny theo nh l Vit ta c: 1 2

1 2

x x m 1

x x 2m 3

+) Khng mt tnh tng qut, gi s A(x1; x

1 + m) v B(x

2; x

2 + m).

x O

1

2

y

I

3 3

2

x + 2

y

+

1 1

y


17 | LOVEBOOK.VN

AOB nhn cos AOB > 0 21 2 1 2 1 2 1 2OA.OB 0 x x x m x m 0 2x x m x x m 0

22 2m 3 m m 1 m 0 3m 6 0 m 2 .

Kt hp vi (**) ta kt lun c cc gi tr m cn tm l m 3

22

; .

Cn nh: AOB nhn OA.OB 0 .

Cu 2.

Nhn xt: Phng trnh dng kh thun, ta bin i tanx = sin x

cos x v quy ng ln th c ngay dng phng

trnh quen thuc vi hng gii l phn tch nhn t chung:

cosx(cos2x + sin2x cosx) (1 sinx)sinx = 0 (*).

n y ta dng my tnh th nghim th thy rng (*) c cc nghim l 0;

4 ;

3

4 ;

2 (sau khi quy ng ta

mi th nghim, ch khng th nghim trc khi quy ng. Bi v nu th nghim trc khi quy ng th c th

lm mt i mt s nghim ca phng trnh, t lm mt i s nh gi khch quan hn v nhn t ca phng

trnh ).

nht l cp nghim i nhau (ta u tin xt trng hp i nhau hoc b nhau, hn km nhau

2 trc), ta

nhn xt:

4 l nghim ca phng trnh

1cos x 0

2

; cn

3

4 l nghim ca phng trnh

1cos x 0

2

. D on rng

1cosx

2

v

1cosx

2

u l nhn t ca phng trnh nhn t chung

ca phng trnh c th l 21 1 1 cos2x

cosx cosx cos x2 22 2

.

Vy ta i theo hng tch nhn t chung cos2x = cos2x sin2x.

(*) cos2x.cosx + 2sinx.cos2x cosx2 sinx + sin2x = 0

cos2x.cosx + sinx(2cos2x 1) (cosx2 sin2x) = 0.

n y th nhn t chung cos2x xut hin ri! Vic d on nhn t ca chng ta thnh cng m mn

Bi gii:

iu kin: x

2 + k (k ) (1).


cos2x + 2sinxcosx cosx (1 sinx).sinx

cos x = 0

cosx.cos2x + 2sinx.cos2x cos2x (1 sinx)sinx = 0

cos2x.cosx + sinx(2cos2x 1) (cos2x sin2x) = 0

cos2x.cosx + sinx.cos2x cos2x = 0

cos2x.(cosx + sinx 1) = 0

k2x k x2x 0 2 4 2

1x x 1 x x k2 x k2

4 22

cos

cos sincos

Kim tra li iu kin (1), ta kt lun c phng trnh c hai h nghim l x =

4 +

k

2 v x = k2 (k ).


LOVEBOOK.VN | 18

Cu 3.

nh hng: Cm gic u tin khi gp phi bt phng trnh ny chc l cng kh ngp . Cha vi ng th,

tm iu kin xc nh ca phng trnh nh .

Khng kh tm c iu kin xc nh ca phng trnh l x 0.

Bc tip theo l bc bin i phng trnh. Mt iu phi tha nhn l bt phng trnh ny kh hc, khi m

ngay trong bc quy ng cng rc ri (mun quy ng ng, phi chia hai trng hp l x > 0 v x < 0), trong

khi li khng nh gi c x nh vo bt phng trnh cho. Khng sao, Nng c m, ma c , cn

gii bt phng trnh iu kin phc tp c phng trnh lo! Tht vy, ta i gii phng trnh tng ng vi

bt phng trnh trn, sau dng bng xt du kt lun nghim ca bt phng trnh.

Bt phng trnh cho tng ng vi:

2 23

23

x 4x 9x 6 x 4x 3x 2 1 1

x 4x 3 1

0

x 2 1

.

Ta i tm nghim ca t s v mu s ca g(x) =

2 23

23

x 4x 9x 6 x 4x 3x 2 1 1

x 4x 3x 2 1 1

v lp bng xt du

ca g(x).

Nghim ca mu s: tm trong iu kin xc nh.

Nghim ca t s l nghim ca phng trnh:

2 23x 4x 9x 6 x 4x 3x 2 1 1 . Trc tin, xin c ph ci v l cc du ngoc phng trnh c d nhn hn:

33 2 3 24x 9x 6x 1 4x 3x 2x 1 (*).

n y chng ta c g? V tri l mt a thc bc ba. V phi l mt cn thc bc 3. Vy gii theo cch thng

thng l lp phng hai v s chng thu c kt qu tt p g. t n ph cng khng kh quan, bi nu t

th ch t c 3 3 2t 4x 3x 2x 1 m khng biu din c lng cn li theo bin t th cng khng n.

Dng nh vic b tc trong cc phng php khc cng vi hnh thc ca phng trnh (mt v bc 3, mt v

cha cn bc 3) gi v p ta i theo phng php dng hm s ny.

Ta s nhm tnh dng hm s bc ba, bng cch thm vo hai v mt lng ng bng lp phng ca v phi (*).

iu ny cng khng c g qu gng p, bi khi cng thm vo hai v mt lng l 3 24x 3x 2x 1 th bn v phi xut hin s hng c ly tha cao nht l 8x3 = (2x)3, l lp phng ca mt lng p.

(*) 33 2 3 2 3 28x 12x 8x 2 4x 3x 2x 1 4x 3x 2x 1 .

Vy hm s ta dng trong bi ton ny l f(t) = t3 + t (l hm ng bin) cn bin i v tri thnh dng (ax

+ b)3 + (ax + b). tm a, b th ta dng phng php h s bt nh:

33 2 3 3 2 2 2 38x 12x 8x 2 ax b ax b a x 3a bx 3ab a x b b

3

2

2

3

a 8

3a b 12 a 2

b 13ab a 8

b b 2

Vic cn li ca l trnh by ra giy na thi nh .

Bi gii:

iu kin: 2 23 x 4x 3x 2 1 1 x 4x 3x 2 0 x 0.


19 | LOVEBOOK.VN

Bt phng trnh cho tng ng vi:

2 23

23

x 4x 9x 6 x 4x 3x 2 1 1

x 4x 3 1

0

x 2 1

(**).

Ta xt du ca v phi bng cch tm nghim ca t s v mu s:

Nghim ca mu s: x = 0.

Nghim ca t s l nghim ca phng trnh:

2 23x 4x 9x 6 x 4x 3x 2 1 1 33 2 3 24x 9x 6x 1 4x 3x 2x 1

33 2 3 2 3 28x 12x 8x 2 4x 3x 2x 1 4x 3x 2x 1

3 33 2 3 22x 1 2x 1 4x 3x 2x 1 4x 3x 2x 1 (1).

Xt hm s f(t) = t3 + t trn . Ta c f (t) = 3t2 + 1 > 0 vi mi t f(t) ng bin trn .

Mt khc (1) c dng 3 33 2 3 2f 2x 1 f 4x 3x 2x 1 2x 1 4x 3x 2x 1

3 3 2 3 2 9 172x 1 4x 3x 2x 1 4x 9x 4x 0 x 0 x

8

.

Lp bng xt du ca v phi (**):

Da vo bng xt du, ta kt lun c tp nghim ca bt phng trnh l:

S = 9 17 9 17

0 08 8

; ; ; .

Bi tp cng c:

1. Gii phng trnh 32 22x x 1 2x 9x 1 11x 1 (p s x = 0 v x = 2).

2. Gii phng trnh 32 3 25x 4x 5x 3 5. 7x 2x 9x 6 (p s x = 1 v x = 8 178

).

3. Gii bt phng trnh 23 23 22x . 6x33x 35 5x 2x 4x 3 (p s

5 97x

12

7 1

9

).

Cu 4.

nh hng: Li mt tch phn bt nh na cha tng hp nhiu loi hm (hm hu t, hm logarit, hm lng

gic). Vi cn khng c g c bit v mu s cha hn hp nhiu hm, nn vic dng tch phn tng phn cng

khng c tc dng g. Tt nhin, nh hng u tin ca chng ta vn l a tch phn v dng: b b

a a

g(x)I f(x)

g(x)

. iu ny cng d nhn ra khi m t s c nhiu s hng tng ng vi mu s, vy nn ta s tch t s thnh

dng f(x).g(x) + g(x) ta s tch nhng du ngoc t s ra, sau tm s hng c cha xlnx v nhm li vi

s hng thch hp, c th l:

x 9 17

8

9 + 17

8 0

0 0 T s VP(**)

Mu s VP(**)

VP(**)

+ +

+

0 0 + + +


LOVEBOOK.VN | 20

T s = sin2x cos x 1 2cos x.x ln x ln x s hng cha xlnx l 2cosx.xlnx nhm c dng f(x).g(x)

(vi g(x) l mu s) th phi nhm (sin2x + 2cosx.xlnx) = 2cosx.(sinx + xlnx).

Lng cn li l (cosx + 1 + lnx) chnh bng o hm ca mu s.

Bi gii:

Ta c:

2

3

sin2x 2cos x.x ln x cos x 1 ln xdx

sin x xI

ln x

3

2

2sin xcos x 2cos x.x ln x sin x x ln xdx

sin x x ln x

3 33 3

2 22 2

sin x x ln xdx 2sin x ln sin x x ln x

sin x x los

n2c x

x

3 2 3 ln 1 ln ln ln2 2 2 3 3

Vy I = 3

2 3 ln 1 ln ln ln2 2 2 3 3

.

Thng tin thm : Dng ton ny tng c xut hin trong thi i hc Khi A nm 2010;

Khi A nm 2011 v trong c thi d b i hc Khi A nm 2012.

Cu 5.

nh hng:

+) Tnh th tch:

u tin phi xc nh c lng tr ng th c cnh bn

vung gc vi mt y CC (ABC).

xc nh c gc gia hai mt phng (ABC) v (CAB)

(c giao tuyn l AB) th ta cn dng mt mt phng vung

gc vi giao tuyn xc nh gc. Thy rng kh thun

li khi c mt cy cu l CC AB, vy nn khng ngi th

m chng ta khng dng thm mt cy cu na l ng

cao CM ca ABC (lu ABC cn ti C nn M l trung

im AB) t bc c mt phng (CCM) l mt

phng vung gc vi AB gc cn xc nh l CMC .

Khai thc c gc th tnh ng cao cc k d dng,

trong khi y xc nh tnh th tch mt cch ngon

lnh nh .

+) Tnh khong cch:

Hai ng thng cn tnh khong cch c mt cnh l cnh y ca lng tr (cnh AB), mt cnh th thuc mt

bn ca lng tr (cnh CB). Li dng tnh cht song song gia cc cnh y (AB // AB), ta tnh khong cch gia

hai ng thng cho nhau bng cch dng mt phng song song, l (CBA) // AB.

Nhim v ca by gi l chn im no trn AB dng ng vung gc n (CBA) cho hp l. Mun thc hin

c iu ny th hy ch rng (CCM) AB, m AB // AB nn (CCM) AB. Vy c mt mt phng i qua mt

C

B

C

B

A

A

M

M

H


21 | LOVEBOOK.VN

im thuc AB (mt phng (CCM) i qua M AB), ng thi mt phng ny cn vung gc vi mt ng thng

trong (CBA) (mt phng (CCM) AB) dng ng cao trong mt phng (CCM) l thun li nht!

Bi gii:

+) Gi M l trung im ca AB. Do ABC cn ti C CM AB. Mt khc AB CC gc gia hai mt phng (ABC)

v (CCM) l CMC = 600.

Ta c: CM = BM.tan CBM = a.tan300 =

a

3.

CC (ABC) CC CM CC = CM.tanCMC = a

3.tan600 = a.

+) Th tch khi lng tr l: VABC.ABC

= CC.SABC

= CC.1

2AB.CM =

31 a a.a.2a.

2 3 3 (vtt).

+) Gi M l trung im ca AB th MM // CC M (CCM).

Ta c: CC AB

CM AB

AB (CCM) nu trong CMM k MH CM (H CM) th AB MH AB MH.

MH (CBA).

+) CMM vung ti M nn 2 2 2 2 2 2

2

a.a

1 1 1 CM.MM a3MH2MH CM MM CM MM a

a3

.

Mt phng (CAB) cha CB v song song vi AB nn:

d(AB, CB) = d(AB, (CAB)) = d(M, (CAB)) = MH = a

2.

Lu : mch trnh by c lu lot th nn l lun v khong cch phn cui cng.

Cu 6.

Trong bi ton ny, chng ta s cp mt phng php khng h mi nhng li t c s dng. l phng

php Nhn vo im cui (Look at the end point). y l mt phng php s gip n gin ha rt nhiu bi

gii, ng thi th n cng l mt trong nhng phng php dn bin m ta t gp.

Phng php ny thng da trn nhn xt n gin sau v hm bc nht:

Gi s f(x) l hm bc nht theo x th:

min{f(a), f(b)} f(x) max{f(a), f(b)} vi mi x [a; b].

iu ny c minh ha mt cch rt trc quan bng th.

Bi gii:

+) Gi s a = max{a, b, c} a b c a b c

b c 1 c a 1 a b 1 b c 1

.

t a b c

P 1 a 1 b 1 cb c 1 c a 1 a b 1

th cn chng minh P 1.

Ta c: (P 1) a b c

1 a 1 b 1 c 1b c 1

.

Xt a b c

f(a) 1 a 1 b 1 c 1b c 1

trn [0; 1]. Theo nh l: (P 1) max{f(0); f(1)}.

Mt khc:

+) f(1) = 0.


LOVEBOOK.VN | 22

+) f(0) =

2 22 2

b cb c 1 bc 1bc b c b c bc 1 2b c

1 b 1 c 0b c 1 b c 1 b c 1

.

max{f(0); f(1)} 0 (P 1) 0 P 1.

ng thc xy ra (a, b, c) = (1; 1; 1), (1; 0; 0), (1; 1; 0) v cc hon v vng.

Cch gii khc:

Gi s a = max{a, b, c}. Khi ta c: a b c a b c

b c 1 c a 1 a b 1 b c 1

.

Nh vy ta ch cn chng minh rng: 1 a

1 a 1 b 1 cb c 1

.

S dng bt ng thc Cauchy ta c:

31 1 1 a 1 a

b c 1 1 b 1 c b c 1 1 b 1 c 1 a 1 b 1 c3 27 27 b c 1 b c 1

.

ng thc xy ra (a, b, c) = (1; 1; 1), (1; 0; 0), (1; 1; 0) v cc hon v vng.

Bi tp cng c:

Cho cc s thc a, b, c, d thuc on [0; 1]. Chng minh rng: 1 a 1 b 1 c 1 d a b c d 1 . Gi : Xem v tri l hm vi bin a dng nh l ln 1 th ta c: f(a) min{f(0), f(1)}.

+) f(1) = 1 + b + c + d 1.

+) f(0) = (1 b)(1 c)(1 d) + b + c + d = g(b).

Tip tc coi y l hm bin b th: g(b) min{g(0), g(1)}.

+) g(1) = 1 + c + d 1.

+) g(0) = (1 c)(1 d) + c + d = 1 + cd 1.

min{g(0), g(1)} g(b) 1 f(0) g(b) 1 min{f(0), f(1)} 1 f(a) 1 (iu phi chng minh).

Cu 7.a.

nh hng: Hnh vung c rt nhiu tnh cht khai thc (tnh cht vung gc; cc cp cnh bng nhau; hai

ng cho ct nhau ti trung im; tnh cht i xng;), vy nn nu gi c ta cc nh ra theo mt s

n t nht th vic x l s khng h kh.

u tin ta im A s vit theo c mt n a. Hai im B v D u c th xc nh ta theo mt n khc,

nhng do im D c mc ni nhiu d kin hn (xD la s nguyn, v I(1; 4) l trung im ca CD u tin

khai thc im D, gi ta D theo mt n biu din c C theo n do bit c th trung im CD) ta

ch dng tt c l hai n cn 2 lin h tm ra c hai n . Hai tnh cht sau s gip ta gii quyt vn

trn

(1) AD ID v (2) trung im ca ng cho AC thuc ng thng BD.

Vi hai mi lin h ny th chc chn s tm c hai n ta A, C, D ta B.

Bi gii:

+) Do A : x y + 1 = 0 A(a; a + 1). Tng t D BD: 5x y 7 = 0 D(d; 5d 7) (d ).

+) I(1; 4) l trung im CD C I D

C I D

x 2x x

y 2x x

C(2 d; 15 5d).

+) ABCD l hnh ch nht nn hai ng cho ct nhau ti trung im mi ng.

trung im Ma d 2 a 5d 16

2 2

; ca AC thuc BD

a d 2 a 5d 16

5. 7 0 4a 20 0 a 52 2

A(5; 6).


23 | LOVEBOOK.VN

+) AD ID 2d 2

AD ID 0 d 5 d 1 5d 13 5d 11 0 26d 126d 14 37d

1

0

3

8

.

(loi)

D(2; 3) C(0; 5) M5 11

2 2

;

B M D

B M D

5x 2x x 2 2

2

11y 2y y 2 3

2

.

.

B(3; 8) (do M l trung im BD).

Vy A(5; 6), B(3; 8), C(0; 5), D(2; 3).

Cu 8.a.

nh hng: u tin xc nh c tm v bn knh ca mt cu (S). Khi c

c bn knh mt cu (S) v bn knh ng trn giao tuyn ca (S) vi (P)

tnh c khong cch t I n (P) nh nh l Pytago. Mt khc (P) li

cha c th gi c dng tng qut ca (P), dng hai iu kin ny l

c th xc nh c phng trnh mt phng (P).

Bi gii:

+) Mt cu (S) c tm I(1; 2; 2) v bn knh R = 5.

Do (P) ct (S) theo mt ng trn c bn knh r = 4 nn khong cch d t

tm I n mt phng (P) l:

d = d(I, (P)) = 2 2 2 2R r 5 4 3 .

+) ng thng i qua im M(0; 0; 5) v c mt vct ch phng l u = (1; 1; 4).

Gi P

n = (a, b, c) l vct php tuyn ca (P) (iu kin a2 + b2 + c2 0). Ta c M M (P) phng trnh

mt phng (P) l: ax + by + c(z + 5) = 0.

Do (P) nn P P

n u n .u 0 a b 4c 0 a 4c b .

+) d(I, (P)) = 3

22 2 2

2 2 2 2 2 2

4c b 2b 3ca 2b 3c3 3 7c b 9 4c b b c

a b c 4c b b c

2 2b 2c

17b 86bc 104c 0 b 2c 17b 52c 0 52cb

17

Nu b = 2c a = 2c chn c = 1 a = b = 2 (P): 2x + 2y + z + 5 = 0.

Nu b = 52c

17 a =

16c

17 chn c = 17 a = 16 v b = 52 (P): 16a + 52b + 17c + 85 = 0.

Cu 9.a.

+) Do khng c trn ha nn xc sut ch Hin thua mt vn l 1 0,4 = 0,6.

+) Gi H, A, B, C ln lt l cc bin c: Ch Hin thng cc, Ch Hin thng cc sau 3 vn, Ch Hin thng

cc sau 4 vn, Ch Hin thng cc sau 5 vn th cc bin c A, B, C xung khc.

+) Khi : H = A B C. p dng quy tc cng xc sut th P(H) = P(A) + P(B) + P(C).

V cuc chi dng li ngay khi c ngi thng vn th 3 nn vn cui cng trong s cc vn chi s l vn ch Hin

thng.

Ta c:

P(A) = 0,43 = 0,064.

Ch Hin thng cc sau 4 vn tc l vn th 4 ch Hin dnh chin thng, v trong 3 trn u tin th: c 1

trn ch Hin thua v 2 trn ch Hin thng.

I

R

r

d


LOVEBOOK.VN | 24

P(B) = C 23

.(0,4)2.0,6.0,4 = 0,1152.

Tng t: P(C) = C 23

.(0,4)2.(0,6)2.0,4 = 0,13824.

Xc sut ch Hin thng l P(H) = 0,31744.

Cu 7.b.

nh hng: Bnh thng, vi mt hnh vung cnh bng 1 chng

hn, ta xc nh c ng v tr cc im M, N c nh trn hnh

vung ri th chc chn mt iu rng, cc gc trong hnh v (bt

k l gc no to t 3 trong 6 im A, B, C, D, M, N trn hnh v) u

c th xc nh c!

Trong bi ton ny th di cnh hnh vung ta cha xc nh c,

nhng cc gc th s khng thay i so vi mt hnh vung c di

bng 1 u nh. bi cho ng thng AN v im M, vy nn

vic i tnh gc MAN s l mt bin php thun li tm c ta

im A, nh vic vit phng trnh AM hp vi ng thng AN

mt gc MAN bit!

Bi gii:

+) t AB = BC = CD = DA = a th BM = a

2 v CN = 2DN =

2a

3.

Dng nh l csin trong MAN ta c:

2 2 2 2 2 22 2 22 2 2 2

AB BM AD DN CM CNAM AN MNcosMAN

2AM.AN 2 AB BM . AD DN

2 2 2 2

2 2

2 2

2 2

a a a 2aa a

2 3 2 3 1

2a a2 a . a

2 3

.

+) A AN: 2x y 3 = 0 A(x; 2x 3) 11 7

AM x 2x2 2

; .

AN c vct ch phng l AN

u = (1; 2).

Ta c: 2

2AN

2 22 2

11 71 x 2 2x

2 2 1 25 85u AM MAN 2 5x 5 5x 25x

2 2211 71 2 x 2x

2 2

. .

cos ; cos

.

x 1 A(1 1)

x 4 A(4 5)

;

;

Vy c hai im A tha mn bi l A1(1; 1) v A

2(4; 5).

Nhn xt, cch gii khc: Bi gii trn ch l mt trong s cc cch c th dng c trong bi ton ny. xc nh

c gc MAN th ta cn c th da vo cng thc cng cung, v d nh:

Cch 1:

A B

C D N

M


25 | LOVEBOOK.VN

1 1BM DN

tanMAB tanNAD 2 3AB ADcot MAN cot MAB NAD tan MAB NAD 12 BM DN 1 11 tanMAB.tanNAD 1 . 1 .

AB AD 2 3

MAN = 450.

Cch 2:

01

2tanMAD tanNAD 3tanMAN tan MAD NAD 1 MAN 45 .

11 tanMAD.tanNAD 1 2.3

V cn nhiu hng na tip cn gc MAN da vo cc nh l sin, cosin, cng cung.

Cu 8.b.

nh hng: (P) i qua hai im cho trc dng gin tip phng php chm mt phng (hai n). Sau da

vo d kin D v C cch u (P) mi quan h t l a : b tm c mt phng (P) xong phim!

Bi gii:

+) Gi phng trnh mt phng (P) l: ax + by + cz + d = 0 (iu kin a2 + b2 + c2 0).

A (P) a + 2b + 3c + d = 0 d = a 2b 3c (1).

B (P) 2a + 3b c + d = 0 c = 2a + 3b + d (2).

T (1) v (2) c = 3a b

4

v d =

5a 11b

4

.

+) Ta c:

d(C, (P)) = d(D, (P)) 2 2 2 2 2 2

b c d 4a 3b 5c db c d 4a 3b 5c d

a b c a b c

3a b 3a bb 4a 3b 5

b c d 4a 3b 5c d 7a 3b4 4

b c d 4a 3b 5c d 3a b 5a 11b 3a b 5a 11b a bb 4a 3b 5

4 4 4 4

.

.

Nu 7a = 3b, chn a = 3 b = 7 c = 4 v d = 23 (P): 3x 7y 4z + 23 = 0.

Nu a = b, chn a = 1 b = 1 c = 1 v d = 4 (P): x y z + 4 = 0.

Nhn xt: Khi bit c mt mt phng i qua hai im th vic dng phng trnh chm mt phng mt cch

gin tip s rt thun li cho vic gii ton.

cng c thm, cc bn hy gii cc bi tp sau:

Bi 1. Trong khng gian vi h ta Oxyz, cho bn im A(1; 1; 1), B(2; 1; 3), C(0; 0; 2) v D(2; 3; 5). Lp

phng trnh mt phng (P) bit (P) i qua hai im A, B, ng thi khong cch t im C n mt phng (P)

gp hai ln khong cch t im D n mt phng (P).

Bi 2. Trong khng gian vi h ta Oxyz, cho bn im A(2; 1; 3), B(1; 2; 3), C(1; 0; 2) v D(2; 2; 1). Lp

phng trnh mt phng (P) bit (P) i qua hai im A, B, ng thi khong cch t im C n mt phng (P)

bng mt na khong cch t im D n mt phng (P).

Cu 9.b.

+) t z = x + yi (vi x, y v x > 0) z x yi .

+) Theo bi ra:

33 3 2 2 3z 12i z x yi 12i x yi x 3xy 3x y y 12 i x yi


LOVEBOOK.VN | 26

2 23 2 3

2 32 3 2 2

x 3y 1 (do x 0)x 3xy x 8y 4y 12 0

3 3y 1 y y 12 y3x y y 12 y x 3y 1

2

2

2 y 1 y 2y 3 0 y 1

x 2x 3y 1 (do x 0)

+) Mun ca s phc z l |z| = 2 2x y 5 .

Nhn xt: Cch t z = x + yi l cch thng c s dng trong cc bi ton v s phc khi cho trc mt

ng thc. Trong bi tp ny, chng ta khng s dng dng lng gic ca s phc bi v s m y cng khng

qu cao, ng thi th trong bi ra cc d kin cng khng xut hin dng tch hay thng p dng dng lng

gic.


27 | LOVEBOOK.VN

S 3


Cu 1 (2,0 im). Cho hm s y = x 2

x 1

(1).

1. Kho st s bin thin v v th ca hm s (1).

2. Chng minh rng vi mi gi tr ca m, ng thng d: y = x + m lun ct th hm s (1) ti hai im

phn bit A, B. Tm m ba im A, B, O to thnh mt tam gic tha mn 1 1

1OA OB

.

Cu 2 (1,0 im). Gii phng trnh (2cosx + 1)(sin2x + 2sinx 2) = 4cos2x 1 (x ).

Cu 3 (1,0 im). Gii h phng trnh

2 2 2

2 3 3

xy x 1 1 3 y 9 3y

3x 1 x y xy 5 4x 3x y 7x 0

(x, y ).


2

0

x 1 sin x cos x cos xdx

x 1 sin x cos x

.

Cu 5 (1,0 im). Cho hnh chp S.ABCD c SA vung gc vi mt phng (ABCD), SA = a, y ABCD l hnh thang

vung ti A v B, AB = BC = a, AD = 2a. Tnh theo a th tch khi chp S.BCD v khong cch t B n mt phng

(SCD).

Cu 6 (1,0 im). Cho x, y, z l cc s thc dng tha mn x z. Tm gi tr nh nht ca biu thc

2

2 2

2z 2y z2x 3zP 2

z xx y y z

.



Cu 7.a (1,0 im). Trong mt phng vi trc ta Oxy, cho hnh ch nht ABCD c AD = 2AB. Gi M, N ln lt

l trung im ca cnh AD, BC. Trn ng thng MN ly im K sao cho N l trung im ca on thng MK. Tm

ta cc nh A, B, C, D, bit rng K(5; 1), phng trnh ng thng cha cnh AC: 2x + y 3 = 0 v im A

c tung dng.

Cu 8.a (1,0 im). Trong khng gian vi h trc ta Oxyz, cho mt phng (P): x 3y + 4z 1 = 0, ng thng

d: x 1 y 1 z

3 1 2

v im A(3; 1; 1). Vit phng trnh ng thng i qua A ct ng thng d v song song

vi mt phng (P).

Cu 9.a (1,0 im). Cho khai trin (1 + 2x)n = a0 + a

1x + a

2x2 + + a

nxn vi n *. Bit a

3 = 2014a

2, tm n.


Cu 7.b (1,0 im). Trong mt phng vi h trc ta Oxy, cho hnh thoi ABCD c ABC = 600. ng trn (C)

c tm I, bn knh bng 2 v tip xc vi tt c cc cnh ca hnh thoi (tip xc vi AB, CD ln lt ti M v N, tung

ca I dng). Bit phng trnh ng thng MN: x + 3 y 1 = 0, ng thng cha cnh AD khng vung

gc vi trc tung v i qua im P(3; 0). Vit phng trnh cc ng thng cha cnh AB, AD.

Cu 8.b (1,0 im). Trong khng gian vi h trc ta Oxyz, cho ng thng : x 1 y 3 z

1 1 4

v im

M(0; 2; 0). Vit phng trnh mt phng (P) i qua im M, song song vi ng thng , ng thi khong cch

gia ng thng v mt phng (P) bng 4.


LOVEBOOK.VN | 28

Cu 9.b (1,0 im). Gii h phng trnh

2

2 2

x 4x y 2 0

x 2 y 0

log log (x, y ).

HT


29 | LOVEBOOK.VN


Cu 1.

1.

Tp xc nh: = \{1}.

S bin thin:

Chiu bin thin:

2

1y

x 1

> 0 vi mi x .

Hm s ng bin trn cc khong (; 1) v (1; +).

Gii hn: x xlim y lim y 1

; x 1lim y

= ;

x 1lim y

= +.

th hm s nhn ng thng x = 1 lm tim cn ng, v nhn ng thng y = 1 lm tim cn ngang.

Bng bin thin:

th:

th (C) ca hm s ct trc tung ti (0; 2), ct trc honh

ti im (2; 0). ng thi (C) nhn giao im ca hai ng

tim cn I(1; 1) lm tm i xng.

2.

nh hng: Vic chng minh d ct th hm s (1) ti hai

im phn bit th kh n gin, ch cn dng phng trnh

honh giao im (l phng trnh bc 2 c hai nghim)

X l iu kin 1 1

1OA OB

. y l mt biu thc i xng ri,

hy th xem nu gi s A(x1; x

1 + m), B(x

2; x

2 + m) (vi x

1,

x2 l nghim ca phng trnh honh giao im

c th dng c nh l Vit) th nh th no nh!

2 22 2

1 1 2 2

1 11

x x m x x m

2 22 21 1 2 2

2 22 21 1 2 2

x x m x x m1

x x m x x m

.

Phi ni rng y l mt biu thc cc phc tp (mc d n

i xng trn c bn th vn dng c nh l Vit, nhng

vic trnh by s rt di), cha k l phi bnh phng hai v

ln mt ln na mi mong xut hin c tng (x1 + x

2) v

tch x1x

2.

1 x O

1

y

I

2

2

1 x O

1

y

I

B

A

d

x + 1

+ + y

+

1 1

y


LOVEBOOK.VN | 30

n y ta nn ngh n tnh cht c bit ca hm s bc nht trn bc nht khi m vic s dng nh l Vit

cho phng trnh honh giao im gp phc tp. th hm s y = ax b

cx d

(vi ad bc v c 0) nhn ng

thng i qua giao im ca hai ng tim cn c h s gc l 1 hoc 1 lm trc i xng (tnh cht ny

cp sch Tuyn tp 90 thi th mn Ton Tp 1). Vi bi ton ny th th (C) ca hm s nhn ng

thng i qua im I(1; 1) v c h s gc k = 1 lm trc i xng v trc i xng ny khng ct th (C). D

vit c phng trnh ng thng : y = x i qua gc ta O. Theo tnh cht i xng ca th th do

ng thng d (c h s gc bng 1) vung gc vi hai im A, B i xng nhau qua . Mt khc O nn

OA = OB. Vy th phc tp ca bi ton c mt phn no ha gii!

Gi ta li c gng tnh OA theo m. Nhn thy rng AB th s tnh c theo m nh nh l Vit, cn d(O, ) hon

ton biu th c theo m. Nh vy dng nh l Pytago s cho ta:

2

22 ABOA d(O )2

, t y tnh c m.

l mt hng lm suy ngh theo bn cht ca vn . khc phc hn ch cch lm trn l l lun hi di

th ta gii bng cch th khn kho nh trong bi gii sau:

Bi gii:

+) Phng trnh honh giao im ca d v th hm s (1) l

x 2

x m x 2 x m x 1x 1

(d thy x = 1 khng l nghim)

2x mx m 2 0 (*).

+) (*) c bit thc = m2 4m + 8 = (m 2)2 + 4 > 0 nn (*) lun c hai nghim phn bit (khc 1) d lun

ct th hm s (1) ti hai im phn bit A, B (pcm).

+) Khng mt tnh tng qut ta gi s A(x1; x

1 + m) v B(x

2; x

2 + m) (vi x

1, x

2 l 2 nghim ca (*)).

Lc ta c: 2 2 2 21 1 2 2 1 1 2 2

x mx m 2 x mx m 2 0 x mx x mx 2 m .

OA = 22 2 2 2 2

1 1 1 1x x m 2x 2mx m 2 2 m m m 2m 4 .

Tng t, ta c 2OB m 2m 4 .

+) 22

m 01 1 21 1 m 2m 4 4

OA OB m 2m 2m 4

Th li, ta thy vi m = 0 th O d; cn m = 2 th O d gi tr m cn tm l m = 2.

Nhn xt: Cch th trn l mt cch th rt c o, ta cn phi nh c c th ng ph vi mi thi tit!

Ngoi ra cc bn cng phi ch vic loi nghim, khng nn xy ra nhng sai lm ng tic l khng loi

nghim.

Cu 2.

nh hng: Ch cn dng hng ng thc th thy ngay 24cos x 1 2cos x 1 2cos x 1 l thy ngay c

nhn t (2cosx + 1). Vic cn li l x phng trnh:

sin2x + 2sinx 2 = 2cosx 1 sin2x + 2sinx 2cosx 1 = 0 (*).

Phng trnh ch c 4 s hng nn chng cn dng my tnh nhm hay th nghim lm g c , ch cn th

nhm vi s hng vi nhau l c. Lu rng: 1 sin2x = (sinx cosx)2 (*) c nhn t l (sinx cosx) ri .

Bi gii:


2cosx 1 sin2x 2sin x 2 2cos x 1 2cos x 1

2cosx 1 sin2x 2sin x 2cos x 1 0


31 | LOVEBOOK.VN

2 22cos x 1 2 sin x cos x sin x cos x 2sin x cos x 0

2cos x 1 sin x cos x 2 sin x cos x 0

21 1 x k2

x x 32 2

x x 1 x k

4

cos cos

sin cosx tan

(k ).

(d thy sinx cosx =

2 sin x4

< 2).

Vy phng trnh c hai h nghim l x = 2

3 + k2 v x =

4 + k (k ).

Cu 3.

nh hng: Nhn cc din chung th thy khng th nh c phng trnh (2) trc c. Hn na phng

trnh (1) cng c dng kh quen, l cha hai biu thc cha cn c kh nng lin hp c, t nhiu lm ta hnh

dung n vic s dng hm s. M mun s dng hm s th phi tch ring x v y ra phi chia hai v cho y2

hai v l hai n tch bit (tt nhin phi xt trng hp y = 0 nu mun chia):

(1) 2 22

3x x 1 1 y 9 y

y

(*).

n y do nng vi dng php chia hai v xut hin dng hm f(t) = 2t t 1 1

nn dn ti mt bi gii

cha cht ch: (*)

2

2 3 3x x 1 x . 1 1y y

.

Vic a y vo trong du cn l cha ng, bi cha xc nh c y m hay dng a vo du cn. Vy nn

cn c thm mt bc nh gi na, bi gii c th c hon thin.

u tin l t iu kin xc nh 2x y 5xy vn cha khai thc c vic chn y > 0. Th nhng ng qun

cn phng trnh (1) cha cc biu thc dng thng gp l 2y 9 y (d chng minh iu ny) VP(1)

dng VT(1) > 0 y 0 v x > 0. Vi x > 0, kt hp vi 2x y 5xy ta suy ra y > 0.

Th ly o hm f (t) > 0 hm ng bin x = 3

y.

Xong vic x l phng trnh th nht! Vic x l tip theo s trong phn Nhn xt trnh b trng lp.

Bi gii:

2 2 2

2 3 3

xy x 1 1 3 y 9 3y

3x 1 x y xy 5 4x 3x y 7x 0

(1)

(2)

+) iu kin: 2x y 5xy .

Ta c: 2 0y 9 y y y VP(1) > 0 VT(1) > 0 y 0 v x > 0 (do 2 2x 1 0y 1

).

+) Lc : (1)

2

2 3 3x x 1 1 1 1y y

(*).

Xt hm s f(t) = 2t t 1 1

trn (0; +).


LOVEBOOK.VN | 32

Ta c: 2

2

2

tf (t) t 1 1 0

t 1

vi mi t > 0 f(t) ng bin trn (0; +).

Mt khc (*) c dng f(x) = 3

fy

(vi x > 0 v 3

y > 0) x =

3

y y =

3

x.

+) Th vo (2) ta c:

3 2 3 23x 1 3x 2 4x 9x 7x 0 3x 1 3x 2 x 4x 12x 8x

2

3 2 2x 3x 2 3x 13x 1 . 4x 12x 8x x 3x 2 x 03x 2 x 3x 2 2

2x 1 y 3

x 3x 2 0 3x 2 y

2

(do x2

3 nn

3x 1x 0

3x 2 2

).

Hai nghim trn u tha mn, vy h phng trnh c hai nghim l (x; y) = (1; 3), 3

22

; .

Nhn xt: Chc chn rng nhiu bn s phn vn cch gii phng trnh (2) sau khi th y vo, ti sao li lin hp

c ngon lnh cnh o nh th. Cch gii trn l cch gii ca nhng ngi thnh tho v c k nng lin

hp, cn by gi chng ta hy i cch gii d nhn hn nh .

Cch gii ny da trn yu t quan trng nht l ta phi nhm c hai nghim ca phng trnh

3 23x 1 3x 2 4x 9x 7x 0 ( l x = 1 v x = 2 c th nhm hoc bm my tnh). Sau dng phng php h s bt nh nh sau:

3 2

3 2 4x 9x 7x3x 1 3x 2 4x 9x 7x 0 3x 23x 1

3 24x 9x 7x

3x 2 ax b ax b3x 1

2 2 2

3 2a x 3 2ab x 2 b 4x 9x 7xax b

3x 13x 2 ax b

(**).

Do chc chn phng trnh c nghim x = 1 v x = 2 nn n c th phn tch dng nhn t:

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

Vy ch cn tm a, b sao cho: 2 2 2 2a x 3 2ab x 2 b k x 3x 2 ( khng nh c nhn t chung ca phng trnh ri th ch cn dng h s bt nh cho mt v thi nh h qu ca nh l Bzu)

22

2

222 2

22

3 3abk a a 1 b 03 2ab 3 a a

3k 3 2ab a 1 b 03 3a2 b 2 a 32 2a2k 2 b aa

7

.......

Ta khng ch a = 3

7

v tnh thm m ca li gii. Khi chn a = 1, b = 0 th ta c:

(**) thnh:

22 3 2 2 3 2 4x x 3x 2x 3x 2 4x 9x 7x x 3x 2 4x 12x 8x

x3x 1 3x 1 3x 13x 2 x 3x 2 x

, y chnh l cch

gii trnh by trong p n!


33 | LOVEBOOK.VN

Khi chn a = 1, b = 0 th thay vo (**) ta thy xut hin c mu s l 3x 2 x , mu s ny bng 0 vi

x = 1 hay x = 2 khng xc nh khng gii theo hng chn a = 1, b = 0.

Vy l thm mt bi phng trnh hay na c gii quyt nh .

cng c vic dng hm v vic dng h s bt nh trong lin hp th cc bn hy cng i lm 2 bi tp sau nh!

Bi tp cng c:

1. Gii phng trnh: 23x 3 3x 1 5x 4 x (p s x = 0 v x = 1).

2. Gii h phng trnh: 2 2

32 3

x 4 x y 1 y 2

6y 5y 1 x 1

(p s (x; y) = (0; 0), (1; 2)).

Cch gii khc cho phng trnh:

3

3 2 33x 1 3x 2 4x 9x 7x 0 3x 2 3 3x 2 x 4x 3x 2 x 0

t a = 3x 2 0 th phng trnh trn tr thnh:

23 2 3x 1

a 3a x 4x a x 0 a x a 2x 1 0 a x 3x 2 xx 2


LOVEBOOK.VN | 34

Phn 2: D on i hc 2014

D on cu 2: Phng trnh lng gic

1. Kin thc cn nh: Cng thc lng gic.

Phng trnh lng gic c bn. Mt s dng ton thng gp. K thut dng my tnh CASIO trong gii phng trnh lng gic. Mt s bin i quen thuc, nhn t thng gp. Cng thc lng gic:

+ Cng thc quy nhn gc:

* Cc gc i nhau:

x x

x x

x x

sin sin

tan tan

cot cot

cos x =cosx * Cc gc b nhau:

x x

x x

x x

cos cos

tan tan

cot cot

sin x =sinx

* Cc gc ph nhau:

x x

2

x x

2

x x

2

x x

2

sin cos

cos sin

tan cot

cot tan

* Cc gc hn km nhau :

sin x sinx

cos x cosx

tan x+ = tanx

cot x+ = cotx

Cch nh: cos i sin b ph cho hn km nhau l tan, cotan. + Cng thc lng gic lin h c bn:

sin2x + cos2x = 1; tanx = sin x

cos x; cotx =

cos x

sin x; 2

2

11 tan x

cos x ; 2

2

11 cot x

sin x .

+ Cng thc cng cung:

cos a b cosacosb sinasinb ; sin a b sinacosb sinbcosa ; tana tanb

tan a b1 tana tanb

Cch nh: tan ca tng bng tng tan chia 1 tr tch cc tan oai hng. + Cng thc nhn: * Nhn i:

2 2

sin2x 2sin xcosx sin x cosx 1 1 sin x cosx

2 2 2 2cos2x cos x sin x 2cos x 1 1 2sin x

* Nhn ba: 3sin3x 3sin x 4sin x ; 3cos3x 4cos x 3cosx .

Cch nh: Nhn ba mt gc bt k || sin th ba bn, cos th bn ba || du tr t gia hai ta || lp phng ch bn, ... th l ok. + Cng thc h bc:

22 2 2

2

1 2x 1 2x x 1 2xx x x

2 2 1 2xx

cos cos sin coscos ; sin ; tan

coscos

3 3sinx sin3xsin x4

; 3

3cos x cos3xcos x

4


35 | LOVEBOOK.VN

Ch : T cng thc h bc hai c th suy ra cng thc h bc bn (cn chng minh trong qu trnh lm bi):

4 4 3 cos4xsin x cos x4

; 6 6

5 3cos4 xsin x cos x

8

.

+ Cng thc biu din theo t = tanx

2:

2 2

2 2 2

2t 1 t 2t 1 tx x x x

2t1 t 1 t 1 t

sin ; cos ; tan ; cot .

+ Cng thc bin i tng thnh tch:

a b a bcosa cosb 2cos cos

2 2

;

a b a bcosa cosb 2sin sin

2 2

a b a bsina sin b 2sin cos

2 2

;

a b a bsina sin b 2cos sin

2 2

Cch nh: cos cng cos bng hai cos cos || cos tr cos bng tr hai sin sin || sin cng sin bng hai sin cos || sin tr sin bng hai cos sin.

h qu thng s dng:

sinx cosx 2 sin x 2 cos x

4 4

;

sinx cosx 2 sin x 2 cos x

4 4

.

+ Cng thc bin i tch thnh tng:

1

cosa cosb cos a b cos a b2

; 1

sinasin b cos a b cos a b2

1

sina cosb sin a b sin a b2

; 1

cosasin b sin a b sin a b2

Ch : Cng thc biu din theo t = tanx

2 khng c trong sch gio khoa, nn nu cn s dng th ta nn chng

minh mt cht (vic ny khng kh nh!).

Cng thc nhn v cng thc h bc thc cht l mt, nn ch cn nh mt trong hai l c. Tng t vi cng thc bin i tng thnh tch v cng bin i tch thnh tng, ta cng ch cn nh mt trong hai l c. Phng trnh lng gic c bn: Cc phng trnh lng gic c bn:

1) x k2

x x k2

sin sin (k ).

2) cosx cos x k2 (k ).

3)

k

x 2x k

tan tan (k, k ).

4) k

x

x

k

a

cot cot (k, k ).

Mt s dng ton thng gp:

+ Phng trnh bc nht vi sinx v cosx c dng: asinx + bcosx = c (trong a2 + b2 > 0).

* Nu a2 + b2 < c2 th phng trnh v nghim.

* Nu a2 + b2 c2 th phng trnh tng ng vi:

2 2 2 2 2 2 2 2

a b c csin x cos x cos x

a b a b a b a b

, trong 2 2

2 2

a

a b

b

a b

sin

cos


LOVEBOOK.VN | 36

+ Phng trnh bc hai vi mt hm lng gic no c dng: at2 + bt + c = 0, trong t c th l sinx, cosx, tanx hoc cotx. Ta cng thng phng trnh bc ba, bc 4 vi mt hm lng gic no . Cch gii th khng c g mi l na.

+ Phng trnh i xng hoc na i xng vi sinx v cosx c dng: f x x x xsin cos , sin cos , thng hay gp

l dng: a sin x cosx bsin xcosx c 0 .

gii ta t t = sinx cosx (iu kin 2 t 2 ) t2 = 1 2sinxcosx a v gii phng trnh n t, sau

tm x.

+ Phng trnh thun nht (ng cp) vi sinx v cosx, dng thng gp nht l thun nht bc hai v thun nht bc ba vi sinx v cosx:

* Bc hai: 2 2asin x bsin xcos x ccos x d 0.

* Bc ba: 3 2 2 3asin x bsin xcosx csin xcos x dcos x esin x f cosx 0.

Cch gii chung: Xt trng hp cosx = 0 (hoc sinx = 0) xem c tha mn khng. Trng hp cosx 0 (hoc

sinx 0) th chia hai v cho cos2x (i vi phng trnh thun nht bc hai) hoc cos3x (i vi phng trnh

thun nht bc ba) (tng t cho sinx) thu c phng trnh vi mt n l t = tanx (hoc t = cotx).

Ch : Trng hp thun nht bc hai vi sinx v cosx th ta nn dng cng thc h bc a phng trnh v phng trnh bc nht vi sin2x v cos2x.

+ Phng trnh dng: asin x bcosx csin y dcosy (vi a2 + b2 = c2 + d2 > 0). Phng trnh ny tng ng

vi: 2 2 2 2 2 2 2 2

a b c dsin x cos x sin y cos y sin x sin y

a b a b c d c d

.

y l phng trnh lng gic c bn.

Ngoi ra cn mt s dng khc ta cp trong cc thi th ca b sch ny, v d nh nh gi, hay nhn thm mt lng, Ni chung nhng dng l ny t kh nng c trong thi.

+ Phng trnh c dng x

f x x x x2

sin , cos , tan , cot , tan = 0 th ta gii bng cch t t = tan

x

2.

+ Phng trnh lng gic khng mu mc: ty bi ton m ta s dng phng php nh gi (thng s dng

nht), thm bt, nhn hoc chia cho mt lng no (p dng vi cc dng c bit), K thut dng my tnh CASIO trong gii phng trnh lng gic: Cc bn c th tham kho phn chuyn sch Ton tp 1 c th hiu r hn v th thut ny, hoc c th ln mng t tm hiu. Mt s bin i quen thuc, nhn t thng gp:

Hin ti th thi i hc c v ra bi phng trnh lng gic kh n gin, v hu nh cc phng trnh u c x l bng phng php nhm nhn t chung. Mt s phn tch nhn t thng dng l:

+) sin2x = (1 cosx)(1 + cosx); cos2x = (1 sinx)(1 + cosx)

+) sin4x cos4x = sin2x cos2x = cos2x.

+) sin4x + cos4x = (sin2x + cos2x)2 2sin2xcos2x = 22 sin 2x

2

.

+) sin6x + cos6x = 1 23sin x

2.

+) 2

1 sin2x sin x cos x ; 2

1 sin2x sin x cos x ;

+) cos2x cosx sin x cosx sin x 2cosx 1 2cosx 1 1 2sin x 1 2sin x

+) 2sin3x sin x 3 4sin x sin x 2cosx 1 2cosx+1 sin x 3cos x sin x 3cosx sin x +) 2cos3x cosx 4cos x 3 cosx 1 2sinx 1 2sinx cosx cosx 3 sinx cosx 3sinx


37 | LOVEBOOK.VN

+)

cosx sin3x cosx cos 3x 2sin x sin 2x cosx sinx cos2x sin2x2 4 4

+) sin x cos3x sin x cos x sin2x cos2x

+) 2sin x cos2x 2sin x sin x 1 1 sin x 2sin x 1

+) 2cos2x cos x 2cos x cos x 1 cos x 1 2cos x 1 Ngoi ra cn c thm cc nhn t cn nh na l +) (1 tanx) cha nhn t (sinx cosx). +) (1 cotx) cha nhn t (sinx cosx).

+) tanx cotx cha nhn t (sinx cosx)(sinx + cosx). Ch : Cn phi s dng linh hot trong vic dng cng thc, v d nh (1 + cosx) th cosx khng c dng nhn

i mt cch cng khai, nhng vn c th dng cng thc nhn i 2x

1 x 22

cos cos ; hay cng thc nhn i

i vi sinx l x x

x 22 2

sin sin cos ;

2. Thng k cng thc s dng trong phng trnh lng gic trong thi i hc: u tin hy im li phng trnh lng gic trong thi i hc qua cc nm t nm 2009:

Cu 1 (Khi A 2009). Gii phng trnh

1 2sin x cosx3.

1 2sin x 1 sin x

Cu 2 (Khi B 2009). Gii phng trnh 3sin x cosxsin2x 3 cos3x 2 cos4x sin x . Cu 3 (Khi D 2009). Gii phng trnh 3 cos5x 2sin3xcos2x sin x 0.

Cu 4 (Khi A 2010). Gii phng trnh

1 sin x cos2x sin x4 1

cos x.1 tan x 2

Cu 5 (Khi B 2010). Gii phng trnh sin2x cos2x cosx 2cos2x sin x 0. Cu 6 (Khi D 2010). Gii phng trnh sin2x cos2x 3sin x cos x 1 0.

Cu 7 (Khi A 2011). Gii phng trnh 2

1 sin2x cos2x2sin xsin2x.

1 cot x

Cu 8 (Khi B 2011). Gii phng trnh sin2xcos x sin xcos x cos2x sin x cos x.

Cu 9 (Khi D 2011). Gii phng trnh sin2x 2cosx sin x 1

0.tan x 3

Cu 10 (Khi A, A1 2012). Gii phng trnh 3 sin2x cos2x 2cos x 1.

Cu 11 (Khi B 2012). Gii phng trnh 2 cosx 3sin x cosx cosx 3sin x 1. Cu 12 (Khi D 2012). Gii phng trnh sin3x cos3x sinx cosx 2cos2x.

Cu 13 (Khi A, A1 2013). Gii phng trnh

1 tan x 2 sin x .

4

Cu 14 (Khi B 2013). Gii phng trnh 2sin5x 2cos x 1. Cu 15 (Khi D 2013). Gii phng trnh sin3x cos2x sin x 0. Sau y s l bng thng k cc cng thc c s dng trong thang im p n ca B Gio Dc.

bi Cng thc c s dng


LOVEBOOK.VN | 38

Quy nhn gc

Lin h c bn

Cng thc cng

Cng thc nhn Cng thc

h bc

Biu din

theo tanx

2

Bin i tng thnh

tch

Bin i tch thnh

tng i Ba

A 2009 X X B 2009 X X D 2009 X X A 2010 X X X B 2010 X D 2010 X A 2011 X X X B 2011 X X D 2011 X

A, A1 2012 X

B 2012 X X D 2012 X X

A, A1 2013 X X

B 2013 X X D 2013 X

Cn v phn loi dng phng trnh lng gic th xin c b qua, bi v ch yu cc bi ton s dng bin i tng ng v phn tch nhn t a v phng trnh lng gic c bn; phng trnh bc nht vi sinx

v cosx; v phng trnh dng: asinx bcosx csiny dcosy (vi a2 + b2 = c2 + d2 > 0).

Nhn xt: Trong cc nm gn y th cng thc c s dng nhiu nht l cng thc cng v cng thc nhn i. y l cc cng thc cc c bn trong qu trnh bin i, nn chc chn l tn sut s dng ca n trong cc

thi s cao. Cc cng thc cn li xut hin vi tn sut s dng nh hn, ring cng thc biu din theo tanx

2

khng xut hin trong thi i hc cc nm gn y na chc l do Chng trnh mi khng cn cp cng thc ny trong sch gio khoa. Ngoi ra ta thy rng ch yu cc phng trnh u c hai dng ch yu, l phn tch nhn t a v phng trnh lng gic c bn hoc phng trnh c dng l:

asinx bcosx csiny dcosy (*) (vi a2 + b2 = c2 + d2 > 0). Phng php nh t n ph th c xut hin trong

khi gii phng trnh, th nhng ta chng cn trnh by vo bi lm v tn thi gian vit t n ph v thay tr li gi tr n ph t tm x. Cn cc loi phng trnh lng gic khng mu mc s dng cc phng php nh gi, hay cc phng php l khc nhng nm gn y khng thy xut hin. C l rng vi vai tr l mt cu cho im, th cu phng trnh lng gic cng ra mc khng qu kh hc sinh c th g im c phn ny. Ring nm 2013 th ra cu lng gic ch n gin c 3 s hng, v vic pht hin ra vn phi ni rng cng qu d dng. D on nm nay s vn l cc phng trnh s dng phn tch nhn t a v cc phng trnh lng gic c bn, hoc phng trnh c dng (*), nhng kh s c nng ln mt cht. Da trn , tc gi xin c d on cc bi sau:

Cu 2A1

. Gii phng trnh

2cos x cos3x 1sin x sin2x.

1 2cos x cos x sin x

Cu 2A2

. Gii phng trnh

2cos x cos x 1

2 1 sin x .sin x cos x

Cu 2A3

. Gii phng trnh 4sin2x 3cos2x = 3(4sinx 1).

Cu 2B1

. Gii phng trnh sin2x + 2cos2x + 4cosx sinx = 1.

Cu 2B2

. Gii phng trnh cotx tanx = cot2x + 1.

Cu 2B3

. Gii phng trnh sin2x cos2x

2.

sin x cos x4 4

Cu 3D1

. Gii phng trnh 2cos3x + cos2x + sinx = 0.


39 | LOVEBOOK.VN

Cu 3D2

. Gii phng trnh cos2x 3sin2x + 9sinx + 6cosx = 8.

Cu 3D3

. Gii phng trnh 24cos x 2 1 cos2x cos3x 6cos x.


LOVEBOOK.VN | 40

D on Cu 3 Phng trnh, bt phng trnh, h phng trnh

1. Chun b kin thc: + Cc phng php gii PT, BPT, HPT nh: S dng phng trnh c bn. Phn tch thnh nhn t.

t n ph. Hm s. Lin hp.

2. Phn dng ton:

3. Phn loi phng php gii: Di y l mt s phng php thng gp trong thi i hc nhng nm gn y:

Dng ton Du hiu V d Phng php thng dng

PT,

BPT

V t Bi ton c cn

:f(x), f(x)3 ...

1. 5x 1 x 1 > 2x 4 (A2005)

2. 235 1 9 2 3 1x x x x

3. 4x 1 + 4x2 1 = 1

4.x3 1

2= 2x + 1

3

BPT c bn Nhn lin hp

nh gi t n ph

Logari

t

M

Bi ton c hm Logarit: ln, log, Bi ton c hm m ax

1. log4(x + 1)2 + 2 = log 2 4 x + log8(4 + x)

3

2. 2 log3(4x 3) + log12

(2x + 3) 2 (A2007)

3. 4x2+x + 21x

2= 2(x+1)

2 1

4. 2x2x + 932x + x2 + 6 = 42x3 + 3xx

2+ 5x

5. 5. 8x + 4. 12x 18x 2. 27x = 0 (A2006)

PT c bn hm s t n ph Phn tch thnh nhn t

HPT

Logari

t

Bi ton c hm m, Logarit: ln, log,

1. {log1

4

(y x) log41

y= 1

x2 + y2 = 25 (A2004)

2. {logx+y(3x + y) + log3x+y(x

2 + 2xy + y2) = 3

4x+y + 2. 4xx+y = 20

3. {x2 + 3x + ln(2x + 1) = y

y2 + 3y + ln(2y + 1) = x

PT c bn

t n ph Hm s

V t Bi ton c cn

f(x), f(x)3 ...

1. {x + x2 2x + 2 = 3y1 + 1

y + y2 2y + 2 = 3y1 + 1 (D b A2007)

2. {x y3 = x y

x + y = x + y + 2 (B2002)

Hm s t n ph

Tng hp

Nhng HPT khng thuc 2 dng trn

1. {x4 + 2x3y + x2y2 = 2x + 9

x2 + 2xy = 6x + 6 (B2008)

2. {x

1

x= y

1

y

2y = x3 + 1 (A2003)

3. {x2 + y+x3y + xy2 + xy =

5

4

x4 + y2 + xy(1 + 2x) = 5

4

(A2008)

Th

Phn tch thnh

nhn t t n ph Hm s

PT, BPT, HPT

tham s

Trong PT, BPT, HPT

c tham s. Thng l cc bi ton bin lun.

1. Tm m HPT c nghim { x + y = 1

x x + yy = 1 3m

(D2004)

Hm s

t n ph Tam thc bc hai

Ta


41 | LOVEBOOK.VN


LOVEBOOK.VN | 42

Phng php

Phn tch phng php V d in hnh

PT,

BPT,

HPT

c bn

PT

V

t

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

f(x) = g(x)2

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

1.7 x2 + x x + 5 = 3 2x x2

BPT

v

t

1.f(x) g(x)

[ {

g(x) < 0f(x) 0

{g(x) 0

f(x) g(x)2

2.f(x) g(x) {

g(x) 0f(x) 0

f(x) g(x)2

3.f(x) g(x) {g(x) 0f(x) g(x)

1.x2 8x + 15 +x2 + 2x 15

4x2 18x + 18

(Dc HN 2000)

Log

arit loga x = n x = a

n 1. log4(x 1) +1

log2x+1 4=1

2+ log2 x + 2

M ax = n x = loga n 2. (x 3)3x25x+2 = (x2 + 6x + 9)x

2+x4

HPT

1. PP th 2. H i xng loi I:

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

vi {f(x, y) = f(y, x)

g(x, y) = g(y, x)

3. H i xng loi II:

{f(x, y) = 0f(y, x) = 0

1. {x3 2xy + 5y = 7

3x2 2x + y = 3

2. {x2 + y2 + xy = 13

x4 + y4 + x2y2 = 91

3.

{

2x + y =3

x2

2y + x =3

y2

Phn tch

thnh nhn

t

Bin i PT, BPT hoc mt PT trong HPT v dng phng trnh tch f(x).g(x) = 0 (hoc f(x, y).g(x, y) = 0). C th: B1: Nhm nghim: Nhm nghim nhn

Nhng PT dng ny thng c nghim p do bc u tin ta nhm nghim. T nghim nhm c ta nh hng nhn t c th xut hin. B2: Phn tch PT theo nhn t d

on.

B3: X l tng phng trnh mi. * Mt s dng thng gp trong thi i hc:

1. PT M, logarit: PT thng phn tch thnh dng sau: (u a)(v b) = 0. 2. HPT c 1 PT l bc 2 vi 1 n (gi s l x) nu xem n cn li (y) l tham s (i khi phi kt hp c 2 PT hoc phi

t n ph vi a c v dng ny): Ta xem y l PT bc 2 vi x. Tnh theo y ri tnh nghim x theo y. T suy ra nhn t chung.

1. 42x+ x+2 + 2x3= 42+ x+2 + 2x

3+4x4 (D2010) Li gii:

Chuyn v:

24x+2 x+2 + 2x3 24+2 x+2 2x

3+4x4 = 0 D nhm c PT c 2 nghim p x=1 v x=2.

Vi x = 1 d thy 24x+2 x+2 = 24+2 x+2 v

2x3= 2x

3+4x4 do ta nhm li v thu c nhn

t chung 24x 24 PT cho tng ng vi:

(24x 24) (2x34 22 x+2) = 0

n y PT d dng gii quyt. + Bi tp tng t D2006

2x2+x 4. 2x

2x 22x + 4 = 0

2. (D 2008)

{xy + x + y = x2 2y2 (1)

x2y y x 1 = 2x 2y (2)

Li gii: Cch 1: Nhm nghim: D thy vi x = y PT (1) lun bng 0. Do (1) c cha nhn t x + y t

d dng phn tch (1) thnh dng (x + y)(x 2y 1) = 0.

T y kt hp vi PT(2) d dng gii HPT.


43 | LOVEBOOK.VN

Sau x l HPT mi thu c khi kt hp vi PT cn li.

Cch 2: Xem (1) l PT bc 2: Cch ny rt hiu qu khi cch 1 nhm nghim gp

kh khn: Trong HPT ny PT (1) l PT bc 2 vi c x v y do chn 1 trong 2 bin l n, ta d dng tnh c

v suy ra 2 nghim x = y

x = 2y + 1

VD: D2012

{xy + x 2 = 0 (1)

2x3 x2y + x2 + y2 2xy y = 0(2)

HPT ny khng d nhm nghim x theo y hoc ngc li. Khi Cch 2 t ra rt hiu qu. D thy (2) l PT bc 2 vi y c dng

y2 (x2 + 2x + 1)y + (2x3 + x2) = 0

= (x2 + 2x + 1)2 4. (2x3 + x2)

= (x2 + 2x + 1)2 Do PT (2) c 2 nghim:

{y = x2

y = 2x + 1

Do PT (2) c ch 2 nhn t (y x2) v (y 2x 1). T :

(2) (y x2)(y 2x 1) = 0 n y x l bc 2 khng kh.

+BT tng t: (A 2011)

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

xy(x2 + y2) + 2 = (x + y)2

t n

ph

PT,

BPT

PP ny c rt nhiu ng dng v lng dng bi v cng phong ph. Do

trong gii hn ca bi vit ti khng th trnh by ht c. Sau y ti xin a ra nhng dng tng qut nht v phng php ny. hiu thm v

phng php ny cc bn tham kho thm trong cc cun sach TUYN TP 90 TON TP 1 VA 2 cng nh cc ngun ti liu khc. 1. t 1 n ph hon ton:

+ Bin i PT v dng f(g(x)) = 0.

+ t g(x) = t + PT tr thnh f(t) = 0. * Mt s dng thng gp : + PT thun nht 2 n:

Bc 2: ax2 + bxy + cy2 = 0

Bc 3: ax3 + bx2y + cxy2 + dy3 = 0

Cch gii: Chia 2 v cho xk. ynk (n l bc

ca PT, k ; 0 k n) (ta thng dng l chia cho xn(k =n)hocyn (k = 0))

+ () + () + ()() = ()

1. 2x2 6x + 4 = 3 x3 + 8

2(x2 2x + 4) 2(x + 2)

= 3(x + 2)(x2 2x + 4)

2. sin3 x + 2 sin x cos 2x 2 cos x cos3 x = 0 3. B2011:


LOVEBOOK.VN | 44

Cch gii: t t=f(x) + g(x) bin i

PT theo t.

2. t 1 n ph khng hon ton:

Dng 1: a PT cho v PT 2 n + Bin i phng trnh v dng:

f(x, g(x))=0.

+ t t = g(x) + PT tr thnh f(x, t) = 0

+ Xem y l phng trnh 1 n (x hoc t), n cn li (t hoc x) l tham s v gii. * Mt s dng thng gp :

() + () = ()()

Trong : u(x) thng bc 1

f(x) thng bc 2 hoc 3 g(x) thng bc 1 hoc 2 Dng 2 : a PT cho v HPT 2 n + Bin i phng trnh v dng:

f(x, g(x)) = 0.

+ t y = g(x) + PT tr thnh f(x, y) = 0. + Ta c HPT (thng l HPT i xng

loi II): {y = g(x)

f(x, y) = 0

+ Gii HPT ny thu c nghim. * Mt s dng thng gp :

+) + =

Cch gii :

t y = bx an

yn + a = bx

PT cho biu th theo t v x : xn + a = by

Ta c HPT {yn + a = bxxn + a = by

y l HPT i xng loi IIHPT c bn

+) + = + + ( ) (*)

Cch gii :

t + = ac2 (y +

d

2c)

Bin i PT v HPT i xng loi 2 vi x v y (HPT c bn)

3. t 2 n ph a v HPT + Bin i PT v dng f(u(x), v(x)) = 0

+ t {u(x) = uv(x) = v

+ chuyn PT v HPT 2 n u v v

3 2 + x 6 2 x + 44 x2 = 10 3x

4. (4x 1) x3 + 1 = 2x3 + 2x + 1 (1)

(1) 2(x3 + 1) (4x 1) x3 + 1 + 2x 1 = 0

t t = x3 + 1 ta c:

2t2 (4x 1)t + 2x 1 = 0

= (4x 1)2 4.2. (2x 1) = (4x 3)2

Do { t =1

2t = 2x 1

n y PT tr nn rt n gin

5. x3 + 1 = 2 2x + 13

D thy t y= 2x + 13

Ta c HPT {x3 + 1 = 2y

y3 + 1 = 2x

6. x22x = 2 2x 1 Bin i PT trn v dng tng qut (*):

x22x = 8x 4

D thy a = 8; b = 4; c = 1; d = 2; e = 0.

Do t + = ac2 (y +

d

2c)

ngha l 8x 4 = 8.13

(y +2

2.1) hay

2x 1 = y 1

hay (y 1)2 = 2x 1 (1)

PT cho biu din theo y v x l

x2 2x = 2(y 1) (x 1)2 = 2y 1 (2)

T (1) v (2) ta c HPT

{(y 1)2 = 2x 1

(x 1)2 = 2y 1

n y ta d dng gi quyt bi ton (HPT c bn)

Bi tp tngt: 3x2 + 6x 3 = x + 7

3

7. 5 x4

+ x 14

= 2


45 | LOVEBOOK.VN

HPT

+ Bin i HPT thnh dng:

{f(u(x, y), v(x, y)) = 0

g(u(x, y), v(x, y)) = 0

+ t {u(x, y) = uv(x, y) = v

+ Chuyn HPT thnh {f(u, v) = 0g(u, v) = 0

+ Tm u, v t suy ra cc nghim (x, y)

1. B2009 : {xy + x + y = 7y

x2y2 + xy + 1 = 13y2

2. D2009: {x(x + y + 1) 3 = 0

(x + y)2 5

x2+ 1 = 0

Hm s

Vi f(x) v g(x) l 2 hm s lin tc trn

ta c: 1. Hm s f(x) n iu trn th: +) f(x) = a c khng qu 1 nghim trn

(nu f() = a th x = ( )

+) f(u) = f(v) u = v u, v .

2. f(x) v g(x) n iu v ngc chiu bin thin trn th f(x) = g(x) c

khng qu 1 nghim trn . 3.

+) f(x) ng bin trn D th f(u) > f(v)

u > v u, v .

+) f(x) nghch bin trn D th f(u) > f(v) u < v u, v D

1. x2 + 3log2 x = xlogx 5

2. 8x3 36x2 + 53x 25 = 3x 53

Ta cn tm cc s sao cho

m(ax + b)3 + (ax + b) = m(3x 5) + 3x 53

ma3x3 + 3ma2bx2 + (3mab2 + a 3m)x

+ mb3 + b + 5m = 3x 53

{

ma3 = 83ma2b = 36

3mab2 + a 3m = 53mb3 + b + 5m = 25

{m = 1a = 2b = 3

Do PT cho tng ng vi

(2x 3)3 + (2x 3) = (3x 5) + 3x 53

(2)

D thy hm s f(t) = t3+t ng bin trn R

Do (2) 2x 3 = 3x 53

n y PT d dng c gii quyt (PT c bn)

3. x + 2. 3log2 x = 3

4. 8x3 + 2x < (x + 2) x + 1

(2x)3 + 2x < ( x + 1)3+ x + 1

Lin hp PP hu ch Gii PT v t (tham kho chi

tit chuyn ny ti quyn 1) 1. (x + 1) x2 2x + 3 = x2 + 1

Tham s

B1: n gin ha PT Dng cc phng php bit nh: Dng PT c bn, t n ph, nhn lin

hp, phn tch thnh nhn t x l PT, BPT, HPT nh bnh thng.

B2: Dng 1 s phng php bin lun

tm m tha m bi ton. *Mt s PPT thng dng

1. Hm s: + c lp tham s m: bin i PT thnh f(t) = g(m) hoc f(t) g(m) hoc

f(t) g(m). + Lp bng bin thin vi f(t). + Bin lun theo yu cu bi ton. VD: +) f(t) = g(m) c nghimmin

Df(t) g(m) max

Df(t)

+)f(t) g(m)c nghim

g(m) maxDf(t).

2. Tam thc bc 2:

Tm m PT 3x2 + 2x + 3 = m(x + 1) x2 + 1 (1)

c nghim. B1: D thy c th dng 2 phng php cp trn gii quyt PT (1) l:

1. t n ph khng hon ton a v HPT 2 n:

t t = x2 + 1 1 x

(1) tr thnh 3t2 m(x + 1)t + 2x = 0 Nu y khng c m th bi ton vn c th gii quyt bng cch xem t hoc x lm tham s. Tuy

nhin y c thm tham s m nn nu lm nh trn s c 2 tham s v 1 n. Bin lun theo hng ny rt phc tp. 2. t n ph hon ton a v PT 1 n:

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

y l 1 dng c th bin i v PT thun nht v ta s dng phng php chia t 1 n ph hon ton (nh phn PP t n ph trnh by). D thy x = 1 khng phi nghim ca PT (1). Do x 1.

Documents

Trích Đoạn Cuốn Toán Tập 2