Giới thiệu phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn toán phần 1

Phucmg phdp vd ki thudt on nhanh thi Dai hoc dat diem cao mon Todn giiip hoc sinh on tap, van dung sang tao kien thuc, tong hop, phan tich, kiem tra nang lye toan dien, hucfng dan chien

PH561P ncuYin P I I U Kiirinii

PHUONO PHAP

D A T D I E M C A O

NGUYEN PHU KHANH

PHUONG PHAP

&

KI T H U A T

DAT DIEM CAO

\SP/ NHA XUAT BAN DAI HOC SlT PHAM

Phucmg phdp vd ki thudt on nhanh thi Dai hoc dat diem cao mon Todn giiip

hoc sinh on tap, van dung sang tao kien thuc, tong hop, phan tich, kiem tra nang lye toan dien, hucfng dan chien thuat giai de thi dai hoc Sach trinh bay noi dung trong tam on thi dai hoc gbm:

- Phucmg trinh, bat phuong trinh, he phucmg trinh dai so He phucmg trinh mil va Idgarit

- Bai todn to'ng hap Bat dang thuc; cue tri cua bieu thuc dai so

- 8 phuang phdp giai phuang trinh lugng gidc nhanh nhat

- Van de lien quan den sophiec, dai sotohffp, xdc sudi

- Cdc bai todn lien quan den ung dung ciia dqo ham va do thi cua ham so: chieu bieh thien cua ham so; cue tri; gid tri Ion nhat va nho nhat cua ham so; tiep tuyen, tiem can (dimg vd ngang) cua do thi ham so; tint tren do thi nhirng diem c6 tinh chat cho truac, luang giao giua hai do thi (mot trong hai do thi Id dubng thang)

- Tim giai han

- Tim nguyen ham, tinh tich phan Ung dung cua tich phan: Tinh dien tich hinh phang, the tich khdi trdn xoay

- Hinh hoc khong gian (tong hap): quan he song song, quan he vuong goc cua dubng thang, mat phdng; dien tich xung quanh cua hinh non trdn xoay, hinh tru trdn xoay; the tich khdi lang tru, khdi chop, khdi non trdn xoay, khdi tru trdn xoay; tinh dien tich mat can vd the tich khdi cdu

- Phuang phdp toa do trong mat phang: Xdc dinh toa do cua diem, vecta Duong trdn, ba dudng conic Viet phuang trinh dudng thdng Tinh gdc; tinh khodng cdch tit diem den dudng thang

- Phuang phdp toa do trong khdng gian: Xdc dinh toa do cua diem, vecta Duong trdn, milt cdu Viet phuang trinh mat phdng, dudng thdng Tinh goc; tinh khoang cdch tie diem den dudng thang, mat phdng; khodng cdch giita hai dubng thdng; vi tri tuang ddi cua dubng thdng, mat phdng vd mat cdu

Mac du tac gia da danh nhieu tam huyet cho cuon sach, nhung sai sot la dieu kho tranh khoi, rat mong nhan duac sy phan bien va gop y quy bau ciia ban doc

de nhung Ian tai ban sau cuon sach dugc hoan thien hon

Tac gia

3

C A U T R U C D E T H I DAi H Q C M O N T O A N '

I PHAN C H U N G (7 diim)

Cau 1 (2 diem):

a) Khao sat su bien thien va ve do thi cua ham so

b) Cac bai toan lien quan den ung dung cua dao ham va do thi cua ham so: chieu bien thien ciia ham so; cue tri; gia trj idn nhat va nho nhat cua ham so; tiep tuyen, tiem can (dung va ngang) cua do thi ham so; tim tren do thi nhiing diem c6 tinh chat cho truac, tuong giao giira hai do thi (mot trong hai do thi la duong thang)

Cau 2 (1 diem):

Cong thiic lugng giac, phuong trinh lugng giac

Cau 3 (1 di£m):

Phuong trinh, bat phuong trinh, he phuong trinh dai so

Cau 4 (1 diem):

- Tim gioi han

- Tim nguyen ham, tinh ti'ch phan

- ung dung ciia tich phan: tinh dien tich hinh phang, the tich khoi tron xoay

Cau 5 (1 diim):

Hinh hoc khong gian (tong hop): quan he song song, quan he vuong goc cua duong thang, mat phang; dien tich xung quanh cua hinh non tron xoay, hinh tru tron xoay; the ti'ch khoi lang tru, khoi chop, khoi non tron xoay, khoi tru tron xoay; tinh dien tich mat cau va the tich khoi cau

Cau 6 (1 diem):

Bai toan tong hop

II PHAN R I E N G (3 diem)

Thi sink chi dugc lam mot trong hai phan (phan A hoac phan B)

A Theo chucmg trinh Chuan:

Cau 7a (1 diem):

Phuong phap toa do trong mat phang:

- Xac dinh tpa dp cua diem, vecto

- Duong tron, elip

- Viet phuong trinh duong thSng

- Tinh goc; tinh khoang each t u diem den duong thing

Cau 8a (1 diim):

Phuong phap toa do trong khong gian:

- Xac djnh toa dp cua diem, vecto

- Duong tron, mat cau

- Tinh goc; tinh khoang each t u diem den duong thang, mat phang; khoang each giiia hai duong thang; vi tri tuong doi cua duong thang, mat phang va mat cau

Cau 9a (

1 diem) :

- S

o phuc

- T

o hop , xa

c suat , thon

g ke

- Ba

t dan

g thiic

; ey

e tr

i cu

a bie

u thii

c da

i so

B Theo chuang trinh

Nang cao:

Cau 7b (1 diim) :

Phuong pha

p tp

a d

o tron

g ma

t phSng :

- Xa

c din

h to

a d

o cu

a diem , vecto

- Duon

g tron , b

a duon

g conic

- Vie

t phuon

g trin

h duon

g thang

- Tin

h goc

; tin

h khoan

g eac

h t

u die

m de

n duon

g th5ng

Cau 8b (1 diim) :

Phuong pha

p tp

a d

o tron

g khon

g gian :

- Xa

c din

h tp

a d

p cu

a diem , vecto

- Duon

g tron , ma

t cau

- Vie

t phuon

g trin

h ma

t phang , duon

g thang

- Tin

h goc

; tin

h khoan

g eac

h t

u die

m de

n duon

g thSng , ma

t phing

; khoan

g eac

h giu

a

hai duon

g thang

; v

i tr

i tuon

g do

i cu

a duon

g thang , ma

t phan

g v

a ma

t cau

Cau 9b (1 diem) :

- S

o phuc

- D

o th

i ha

m pha

n thu

e hii

u t

i dan

g y = +

bx +

e quan

- S

u tie

p xii

e cu

a ha

i duon

g cong

- H

e phuon

g trin

h m

u v

a iogarit

- T

o hop , xa

c suat , thon

g ke

- Ba

t dan

g thiie Cu

e tr

i eii

a bie

u thii

c da

i so

PHlTdNG TRINH , BA

T PHlTdN

G TRINH , H

E

I

P HlT

O fNG TRINH

Phuong trinh, bat phuong

trinh, he phuong trinh

dai so.

H( phuong trinh mil

va Iogarit

MO

T S

O H

E P HU ON

G T RI NH C

O B

AN

1) H

e ba

c nha

t ha

i an , b

a a

n

2) H

e gb

m mo

t phuon

g trin

h ba

c nha

t v

a phuon

g trin

h ba

c ca

o

• Phuon

g pha

p ehung : S

u dun

g phuon

g pha

p th

e

- H

e ha

i phuon

g trin

h

- H

e b

a phuon

g trin

h

3) He doi xung loai 1

Phuong phap chung: Dat an phu a = x + y; b = xy

4) He doi xung loai 2

Phuong phap chung: Tru tung vehai phuong trinh da cho nhau ta dugc: (x - y).f(x; y) = 0

5) He phuong trinh dang cap bac hai

Xet truong hop y = 0 Vol y ?t 0, ta c6 the tien hanh theo cac each sau:

- Dat an phu y = t.x

- Chia ca hai ve'cho y^, va dat t = —

y

M O T SO P H U O N G P H A P G I A I H E P H U O N G T R I N H

1) Phuong phap the

• Phuong phap: Ta riit mot an (hay mot bieu thuc) t u mot phuong trinh trong he va

the vao phuong trinh con lai

• Nhqn dang: Phuong phap nay thuong hay su dung khi trong he c6 mot phuong

trinh la bac nha't doi voi mot an nao do

2) Phuong phap cong dai so

3) Phuong phap bien doi thanh tich

4) Phuong phap dat an phu

5) Phuong phap ham so

6) Phuong phap su dung bat dang thuc

C/iM y: ting dung dao ham gidi todn phuong trinh, he phuong trinh

Su dung cac tinh chat ciia ham so de giai phuong trinh la dang toan kha quen thupc

Ta thuong c6 ba huong ap dung sau day:

Huang 1: Thuc hien theo cac buoc:

Buac 1: Chuyen phuong trinh ve dang: f(x) = k

Buac 2: Xet ham so y = f(x)

Buac 3: Nhan xet:

• Voi x = Xg <=> f(x) - f(xg) = k, do do Xg la nghiem

• Voi X > Xg o f(x) > f(xQ) = k, do do phuong trinh v6 nghiem

• Voi X < Xg <=> f(x) < f(xQ) = k, do do phuong trinh v6 nghiem

• Vay Xg la nghiem duy nhat aia phuong trinh

Huong 2: Thuc hien theo cac buoc:

Suae 1: Chuyen phuong trinh ve dang: f(x) = g(x)

Buoc 2: Dung lap luan khang dinh rang f(x) va g(x) c6 nhung tinh chat trai ngugc

nhau va xac dinh Xg sao cho f ( X g ) = g(xg)

Buac 3: Vay Xg la nghiem duy nhat ciia phuong trinh

Huang 3:

Thuc hie

n the

o ca

c buac :

Buac 1:

Chuyen phuon

g trin

h v

e dan

g f(u ) = f(

v)

Bwac 2:

Xet ha

m s

o y = f(x) , dun

g la

p lua

n khan

g din

h ha

m s

o do

n dieu

Buac 3:

Khi d

o f(u ) = f(v) o

u = v

Cac v

i du

|

Vi d

u 1 Gia

i ca

c phuon

g trin

h sa

u tre

n ta

p s

o thuc :

1 V5 x

-l

-7 3x

+ 1

3 2 +

- =-2x

-4 + - x X 3 V

3 ~ + X = x

+ X XV 2x

4 2'"'+2^''-9.2'''+18

3 = 2^''*'+2''"+9.2^*"

Lcri giai

1 Die

u kien : x > —

5 a ch h d Phuong trin

o tuon

g duon

g vo

i

VSx

1

V3x

+ 1

3 =

-if VS x

1

V3x +13)(

VSx

-1 +

V3x

+13)

Truang hap

1: V

Sx l

%/3x +

13 =

0 o

V sx -l = V sx TlS

, bin

h phuon

g ha

i v

e ro

i rii

t

gon t

a dug

c x = 7

(thoa man)

Trucrng hap

2: V

Sx -l + V3x

+ 1

3 =

6 (1 )

Neu x>

l th

i V

T (l ) >

>/4 + Vl

6 =

, co

n ne

u x<

l th

i V

T (l ) <

^/4 + Vl6 = 6

De tha

y x

= 1

la nghie

m phuon

g trin

h (1)

Vay phuon

g trin

h ba

n da

u c

6 ha

i nghie

m x = 1 , x = 7

2, Die

u kien : 1 +

- ^

0 o > 0

o x

<

-2 hoa

c x

> 0 x X

Khi d

o phuon

g trin

h d

a ch

o vie

t lai

: x

^l +

- = 2x^

4x

+ 3

+ Vd

i X > 0 , phuon

g trin

h

o>

yx 2

+2

x =-2

(x 2

+2x) +

Dat: t

= V x^

+2 x,t

>0 , t

a c

6 2t

^ +

1-3 =

0 o

t =

1 hoa

c t = , doi chie

u die

u kie

n

ta dug

c t

= l , tu

c la phai c

6 Vx

^ +2

x =

1 o x^ + 2x =

1, phuon

g trin

h na

y c

6 nghie

m

x = -l + \f2 tho

a ma

n die

u kien

+ Vd

i X ^

-2, phuon

g trin

h o

-V x^

+2x = -2(x

^ + 2x) +

3

Dat t = >

/x

^+

2x ,t>

0, t

a c6 : 2t

^ -

1-3 =

0 o

t = | hoa

c t

=

-l, do

i chie

u die

u kie

n

ta dug

c t = ^ , tu

c l

a pha

i c

6 4(x

^ + 2x) =

9 <=

> 4x

^ + 8x

9 =

, phuon

g trin

h na

y c

6

nghiem

x = ~^

'^'^''^ man

dieu kien

Vay phuong trinh da cho c6 nghiem x = - 4 - V 5 2 ; X = - 1 + r/2

3.Dieu kien: x # 0 , x - i > 0 , 2 x - - > 0

X X Phuong trinh da cho dugc bien doi ve dang: x - — = Ix - — - p x - — (1)

Ap dung cong thuc: — - j = , vdi a > 0; b ^ 0; a ^ b va Va + Vb > 0

V a - v b

- X Khi do ( l ) o x — =

X

^ 4 4

* Neu X — > 0 thi — x < 0, khi do tu (2) ta c6 ve trai Ion hon 0, ve phai be hon 0, v6 li

4 4

* Neu X — < 0 thi — x > 0, khi do tu (2) ta c6 ve trai be hon 0, ve phai Ion hon 0, v6 li

X X

Vay X — = 0 o x - 4 = 0=>x = 2 (thoa man dieu kien)

X Phuong trinh da cho c6 nghiem duy nhat x = 2

4 Phuong trinh da cho tuong duong v6i

(2'" +2 -2(2" +2 -72(2" +2 ") + 185 = 0

o ( 2 " + 2 " ) ' - 2 - 2 ( 2 " + 2 - " ) ' - 7 2 ( 2 " + 2 ") + 185 = 0

<^(2" +2 -6(2" +2 - 7 2 ( 2 " + 2 ") + 189 = 0

Dat t = 2" + 2 ", t > 2, phuong trinh da cho tro thanh: t^ - 6t^ - 72t +189 = 0

c = > ( t - 3 ) ' ( t ' + 6 t + 2l) = 0 o t = 3, vi t ' + 6 t + 21 > 0 , V t > 0

Voi t = 3 o 2 " + 2 - = 3 « 2 " = ^ ^ « x = l o g / ^ * ^ ^

Vay phuong trinh da cho c6 nghiem duy nhat x = log^

/ /—

3±V5

V i du 2 Giai cac phuong trinh sau tren tap so thuc:

1 log25(x^-8x + 15)^ = | l o g ^ ^ + log5|x-5

2

2 log, (4 - x ) ' + ^ log, (x + 2)' = 3 + log, (x + 6y

3 4" (V2.6"-4" + V4.24" -3.16") = 27" -12" + 2.8"

gidi

1 Die

u kien : x > 1;

x # 3;x #

5 Phuon

g trin

h d

a ch

o tuon

g duon

g vo

i

logs

x'' -8x +

15

= ' 0g 5^

+ l0g

5 X

-5

X-1 -y g5 lo = -5 5 X 3 +log x- g og ol -+

log5|x-5

o2

|x -3 =

x-l<:=>2x-6 =

-l hoa

c 2 x-6 = l -x <:=>x =

- (v

i x^5) ' 3 ^

7 = m x 6 nghie h c g trin Vay phuon

- 3

2 Die

u kien :

-6 <

x <

4 v

a x ;t

-2 (*

)

I^huong trin

h d

a ch

o

< => 31og^(4-x)-31og

^ x + 2 = 3-31og

^ {

x + 6)

o log

^ (4 ~ x ) + iog^ (

x + 6) =

1 + log^ |

x + 2|

o (4 x)(x + 6) = 4|x +

2

<=>4(x + 2) = ( 4-x) (x + 6 ) hoa

c 4(

x + 2)

=-(4 -x )(

x + 6) (v

i (*

) ne

n ( 4-x)

(x

+ 6

)

>0

)

+ V6

i x -+

6x -1

6 = 0c:>x = 2,

x =

-8

+ V6

i x' -2 x-32 = 0<=>

x =

l + 733,x = l -V 33

Vay phuon

g trin

h c

6 ha

i nghie

m x = 2

; x = 1

- V3

3

3 Die

u kien : 2 6"

- 4

" >

0 v

a 4.24

" 3.16" >

0 «

x >

log,

-Ta C

O ca

c dan

h gi

a sau : 4"^2 6"

-4

" = 2".2"^2.6

" -4

" <

^ 2" (4

" + 2.6

" -4

") = 12"

4" V4.24"-3.16

" = 4" ^^4

" (4.6

" -3.4") <

^4" (2

" + V4.6

"-3.4")

= -8" +i2

\2"V4.6

"-3.4"

<-8

"

+-2

"

(4

"

+4.6

"

-3.4") = 12" ^ ' 2 4 2 2

Do d

o 4

" (V2.6

" -4

" + ^4.24"-3.16") <

2.12", dSn

g thu

c xa

y r

a kh

i x = 0

27"-12"+2.8" = 27" +8

" -12

" >

3N

/27".8".8" -12

" = 2.12", da

u ban

g xa

y r

a kh

i x

= 0

Vay phuon

g trin

h na

y pha

i c

6 nghie

m x = 0

Vi d

u 3 Gia

i ca

c phuon

g trin

h sa

u tre

n ta

p s

o thuc :

1.2x^+6x^+6x + l =

3| x

+ 2

2 8

log4 N/X^

9 +

3^2 log4

(x + 3)^ =

10 +

log2

(x

3)^

Lai gidi

l.D

at 3

|^i±?.=y + i c^

^i

±i =

y 3+

3y 2+

3y + i ox = 2 y^

+6 y^

+6

y (1)

Theo de bai ta CO p h u a n g t r i n h 2x + 6 x + 6 x + l = y + 1 <=> y = 2x + 6 x + 6 x ( 2 ) ,

Ttr (1) va (2) s u y ra

2x^ + 6x2 + 6x - (2y^ + 6y2 + 6 y ) = y - X o 2(x^ - y^) + 6(x2 - y^) + 7(x - y ) = 0

o (x - y)(2x2 + 2y2 + 2xy + 6x + 6y + 7) = 0

o x = y hoac 2x^ + 2 y 2 + 2 x y + 6x + 6 y + 7 = 0 (3)

P h u a n g t r i n h (3) v6 n g h i e m v i

= (x + y + 2 ) 2 + ( x + l ) 2 + ( y + l ) 2 + l > l V x , y

^i-#-Vay x = y , t h a y vao (2) ta d u o c

2x'' + 6x2 + 6x = x o + 6x2 + 5x = 0 o ^^2x2 + 6x + 5) = 0 C5> x = 0

V a y X = 0 la n g h i e m p h u o n g t r i n h

2 D i e u k i e n : x < - 3 hoac x > 3

81og4 7x2 - 9 + 3 j 2 1 o g 4 ( x + 3)2 = 10 + \og^(x -sf

2log2(x2 - 9 ) + 3 j l o g 2 ( x + 3)2 = 10 + l o g 2 ( x - 3 ) 2

<=>

log2(x + 3)2 + 3^/log2(x + 3)2 - 1 0 = 0

D a t t = 7 l o g 2 ( x + 3)2 > 0 Ta CO p h u a n g t r i n h :

t ^ O

t 2 + 3 t - 1 0 = 0 o t = 2 o l o g 2 ( x + 3)2 = 4 o (x + 3)2 = 16 <=> X = - 7 hoac x 1

D o i c h i e i i d i e u k i e n , ta c6 n g h i e m p h u a n g t r i n h la x = - 7

V i d u 4 G i a i cac p h u a n g t r i n h sau t r e n t a p so t h u c :

1 x^ + 3x2 + 9x + 7 + ( x - i o ) V 4 - x = 0 2 V x ^ + yjx^ +x^ +x + \ 1 + Vx'* - 1

Lai giai

1 D i e u k i e n : x < 4

Bien d o i p h u a n g t r i n h ve: (x +1)"' + 6(x +1) = V 4 - x ( 4 - x + 6)

Dat u = x + l , v = \ / 4 - x ^ 0 T a c 6 p h u a n g t r i n h : u ' ' + 6 u = v'' + 6 v

o ( u ' ' - v^) + 6 ( u - v ) = 0 <=> u = V hoac u2 + u v + v2 + 6 = 0

Trudmg hQfp 1: u2 + u v + v2 + 6 = 0 <=> ( u + —)2 + + 6 = 0 , v 6 n g h i e m

Trumtghgp 2: u = v <=> V 4 - x = x + 1 <=>

2 4

x + 1 ^ 0

4 - x = x 2 + 2 x + l <=> X =

2 Die

u kien : x > 1

Dat a = V x-l,b = Vx

^

+x-^ +

X +

1, vo

i a>0,b>0,tac

6

Vx '* -l=

\/(

x-l )(x

^+

x2

+x +

= V^^.Vx

^ + x^ +

X +

1 = a.b

Phuong trin

h d

a ch

o tr

o thanh :

a +

b =

l +

b<

»(a -l) (b -l) =

0<=>a-l =

0 hoa

c b-l=0<::>

a =

l hoa

c b

= l

+ Vd

i a

=lt

hi Vx -1 = l<

:i

x-l = l<=>

x =

2

+ V6

i b = l th

i Vx^+x^+

x +

1 = l»

x'' +x

^+

x = Oc=>x^x^+x +

j = 0<=>x = 0(loai,do

X >

1) hoa

c x

+ X

+1 =

0 (phuon

g trin

h na

y v

6 nghie

m v

i x + x +1 =

moi x )

Vay phuon

g trin

h d

a ch

o c

6 nghie

m du

y nha

t x = 2

Vi d

u 5

Gia

i ca

c phuon

g trin

h sa

u tre

n ta

p s

o thuc :

1 X

+ V

4"

X^ =

2 +

3xv'4-X

V

2,

\2

+ - >

0, vo

2 2x'-6

x + 10-5(x-2)Vx +

l

=0

3 4(

N/X +T

-3

)X

' +

(I 3V

X +

1 -8

)X

~4

VX

-1

3

-0

1 Die

u kien :

-2 <

x <

2

Dat t = x + \/4-x ' =

> t

" =

4 +

2x\l4~x- =>

wli-x' = ~ ^,

2

Phuong trin

h d

a ch

o tr

o thanh : t = 2 + 3 -—

^c

>3 t'-2t -8 = 0ci>

t = hoac t

= 2

+ Vo

i t = 2 , t

a c6 : x + V 4-x' =

2 <=

> ^Ji-x'

=

2-x o

4 /

4 /

4

+ Vo

i t = — , t

a C

O x

+v4

-x^ = — <

= > v4

-x^ =

2-x>

0

4~x^ =4-4

x + x^

o

x =

0

x =

2'

<=>

<

x

-i

3 9x'+12x-10 =

0 x<

-i

3

x = -2±Vl4

=> X =

3

-2 -V l4

Vay phuon

g trin

h d

a ch

o c

6 b

a nghie

m x = 0

; x = 2

; x -2-Vi4

2 Die

u kien : x >

-1

Phuong trin

h d

a ch

o tuon

g duon

g 2(

x ~ 2)" + 2(x +1 ) 5(x 2)>/x +

l =

0