4 {Cet hop voi dieu kien suy ra tap nghiem ba't phuang trinh da cho la

Một phần của tài liệu phân loại và phương pháp giải đại số 10 (Trang 174 - 178)

S = 0;1

4 u[4; + 0 3 )

pKXD: 1 - X > 0 <=> -1 < X < 1 ^ • V w , 1

1 3 x^ X

Bat phuong trinh <=> ^ -1 + 2 > , <=> ^ - 3 , _r + 2 > 0 . 1-x^ 7177 i - x 2 7 ^

Dat t =

t < l c=>

ta CO bat phuang trinh : t^ - 3t + 2 > 0 <=>

V-'

< l o V l - x ^ >x<=>

-1 < x < 0

0<x<l <^

t < l t > 2

l - x 2 > x 2

-i<x<o < * i' 3 ' rV

l o - l< x < - ^ .

t>2<=> • 2 o X > 2Vl7x^ ô • 0< x < l 2 x 2> 4 ( l- x 2 ) V 5

Vgy nghiem ciia bat phuong trinh da cho la: T = f I 1 ^ ' ' V 2 ; r 2 ]

Vi du 6: Cho phuong trinh %/x + Vl - x + Vx - x^ = m a) Tim m de phuang trinh c6 nghiem duy nhat j) Tim m de phuang trinh c6 nghiem.

Lot gmi

DKXD: 0 < X < 1 ' Gia su phuong trinh c6 nghiem duy nhat XQ tuc la ta c6 ' '

r — I I , r

v'^o + ~ '^o + V'^o ~ '^o) = t3 CO the viet lai la "

V ^ ~ ^ + sf^) + 7 ( 1 " = "1 do do 1 - X ( , cung la nghiem ciia phuong

Wnh da cho , ...

, (ằ>•.' -

'^o do phuong trinh c6 nghiem duy nhat thi xg = 1 - Xg <=> xg = - thay vao

CO m = V o i m J

I + 2V2 2

I + 2V2 r — / 2" 1 + 2V2

Vx + Vl - X + vx - X = ( )

^ ta CO phuong trinh Vx + Vl - x + Vx - x"^ = (*)

Phiin loai va pliumtg plidp s'"' •'<'>'<>_

, I J r~r-. c X + 1 - X 1 A p d u n g BDT cosi ta co Vx - x^ = ^ x ( l - x) < ^ = - Mat khac (V^ + V T ^ f = l + 2 7 x( l - x) < 2= > V ^ + 7 T^ < > / 2 Suy ra ^yx + ^/^ - x + V x- x ^ < , dang thuc xay ra <=> x = i Do do p h u o n g trinh (*) co nghiem duy nha't ^ . ^ ^ ^ ^ Vay m = — la gia tri can t i m . „ , j/ ' I ^j^,,^ ^ b) Dat t = N/^ + V T ^ ^ t^ = 1 + 2^ x( l - x )

Theo cau a ta co 1 < (Vx + ^l-xf = 1 + 2^ x( l - x ) < 2 S J S u y r a l < t < x/ 2 , j;-. j • j , , • , ^ Phuong trinh tro thanh t t ^ - l = m ằ t 2 + 2 t - l = 2m (**)

Phuong trinh da cho co nghiem khi va chi khi phuong trinh (**) co nghiem thoa man l < t< N/ 2

<=> D o thi ham so y = t^ + 2t -1 tren [ l ; V2 ] cat d u o n g thMng y = 2m . X e t h a m s o y = t 2 + 2 t - l tren \v,^'^ ^Mtmm^

i"'

t 1 ^

i"' y _ _ _ _ _ ^ l + 2V2

0 —

J, Suy ra p h u o n g trinh da cho co nghiem <=> 1 < 2 m < 1 + 2^

, 1 ^ ^1 + 2V2 hay <=> - < m <

' 2 2 i'l'"- •'!!•;

S " ' 3 : P h u o n g phap danh gia

D o i v o i p h u o n g trinh ta thuong lam n h u sau \

< Cach 1: T i m mot nghiem va chung m i n h no la nghiem d u y nhat.

Cach 2: Bien doi hSng dSng thuc dua ve bat p h u o n g trinh f (x) = 0 tron^

f (x) la tong cac b i n h phuong. ^ Cach 3: V o i p h u o n g trinh f(x) = g(x) co tap xac d j n h D

Neu f ( x ) > m ( x )

g ( x) < m ( x ) , Vx e D t h i f(x) = g(x) o f(x) = m(x) g(x) = m(x)

yj{ 7- Giai cac p h u o n g trinh sau

J j ^ + ^/jr^ = 6 b) +75774= ^ / 2 ^ + ^/3^

^ 3) , .

'^^^ • c Lcrigidi

3) DKXD: x < 2 . ,1,,,.;;,.,,; o . ' . „ , , r . .

Ta thay rang p h u o n g trinh co mot nghiem la x = - va ta chung m i n h do la nghiem duy n h a t . \l;f'"*' ^"v':'i V j ;i<)7 g r u w b g m n i l ônrn

Thgtvay ; ,|, • ; 7 r • ^ / f 7 ' -

* Voi X < - ta C O > 4 => / > 2 va

2 3 - x V 3 - X V 2 - X >

2- 3 , 2

= 4

• + > 6 =:> phuong trinh v6 nghiem.

• Voi ^ < X < 2 ta C O — ^ < 4 => . / — ^ < 2 va ằ =4 2 - 3 2 3 - x ^ ' - ^ v 3 - x - ^-"vri^ 2 Suy ^/jr~ ^/^r^ ^ ^ ^ phuong trinh v6 nghiem.

V|iy p h u o n g trinh co nghiem duy nha't x = - . , , 1 2 > / c,>""'0••;••(!•,- x|x

b ) D K X D : x > 0 x.*:'=xi;i Phuong trinh tuong d u o n g voi V2x + 3 - Vx + 4 = f x + 8 - 7 3 x

De thay X = 1 la nghiem ciia phuong trinh

Voi x > 1 ta C O 2x + 3 > X + 4 o V2x + 3 - N/X + 4 > 0

Va ( x- l ) ( 9 x + 8 ) > 0 o 9 x 2 - x - 8 > 0 o x + 8< 9 x 2 ô . f ^ - V 3 ^ < 0 Suy ra p h u o n g trinh v6 nghiem

Voi 0 < X < 1 ta CO 2x + 3 < X + 4 ô . V2x + 3 - V x + l < 0

Va ( x - l ) ( 9 x + 8 ) > 0 o 9 x 2 - x - 8 > 0 <::>x + 8> 9 x 2 c > f^ - 7 3^ > 0 Suy ra p h u o n g trinh v6 nghiem

Ygj^phuong trinh co nghiem duy nhat x = 1 . 8: Giai cac p h u o n g trinh sau

^ - 9 x + 2 8 ^ 4 V ^ b ) _ x^ 2 2 x+ V l± 2 ^ = 2- x 2 Laigidi

®KXD: x > l ' ' , . , .... • .

* nuong trinh tuong d u o n g voi

'ô^-10x + 25 + ( x - l ) - 4 V x ^ + 4 - 0 <:^(x-5)^+(7x71-2)^=0 (*) 349

V i (x - 5)^ + (N/X - 1 - 2)^ > 0 v o i moi x nen Phuang trinh (*) x- 5 = 0

<=> X = 5 Vay p h u o n g trinh c6 nghiem duy nha't x = 5 .

f l - 2 x > 0 1 1

b) D K X D : ^ ^ ^ ô - - < x < - M r .

Phuong trinh t u o n g d u o n g v o i ( > /T^ ^ + V l + 2 x ) = (2 - x^)

2

2 + 2 V l- 4 x ^ = 4 - 4 x ^ + x'* o ( V l - 4 x ^ - 1

x = 0

; <r>x = 0

V i- 4 x ^ - 1 = 0 . .' ; •' -;'^r: • M

+ x'* = 0 y - (. / X -

Vay p h u o n g trinh c6 nghiem d u y nhat x = 0.

V i d u 9: Giai cac p h u o n g trinh sau 2x^ + X- 1

1 + 3VX + 1 b) \/x^ - 1 + X = Vx^^ - 2

a) Gia s u p h u o n g trinh c6 nghiem, k h i do nghiem cua no phai thoa man x + l > 0

x ( x - l ) > 0 C : > X 6 { - 1 } U [ 1 ; + M ) 2x^ + x - l > 0

R6 rang x = - I khong la nghiem cua p h u o n g trinh, ta xet x > I Phuong trinh da cho o 2x^ + x - 1 = Vx^ - x + 3^Jx{x^ -1) (*)

Ap d u n g BDT cosi ta c6: Vx^ - x < 3 7 x ( x ^ - 1 ) <

3(x + x 2- l )

Suy ra VP(*)

3(x + x ^ - l ] ,

• * ) < - - + - ^ ^ = 2 x 2 + x - l = VT(*) 1 ± V 5 Dang thuc xay ra k h i va chi k h i x - x - 1 = 0 ằ x =

T h u lai p h u o n g trinh ta tha'y x = llJl. la nghiem cua p h u o n g trinh Vay p h u o n g trinh c6 nghiem duy nha't x = 1 + 75

350

D K X D : x 3 - 2 > 0 < = > x > ^ Gia s u p h u o n g trinh c6 nghiem

Su d u n g bat dMng thuc cosi, ta duoc ^x^ - 1 < 2 ( x - l ) + (x + l ) + 4 ^ x + l_

6 2 pCet hop v o i p h u o n g trinh suy ra + x ^ Vx"' - 2

^ 4( x ^ - 2) < ( 3 x + l)2 ô ( x- 3 ) ( 4 x 2 + 3 x + 3) < 0 c ^ x< 3 ! • : N h u vay ta c6 ^ < x < 3. (**)

Ta CO ^^x^ - 1 > X - 1 o X + 1 > (x - 1 ) 2 ô . x(3 - x) > 0 (dung v o i d k (**)) va Vx^ - 2 < 2x - 1 c=> (x -3){x^ - x + 1) < 0 (dung v o i d k (**))

Suy ra ^x^ - 1 + x > 2x - 1 > Vx^ - 2 . . .

Dang thuc xay ra k h i x = 3. T h u lai ta tha'y x = 3 la nghiem ciia p h u o n g trinh da cho

Vay p h u o n g trinh c6 nghiem duy nhat x = 3.

Nhan xet: V o i dieu ki^n xac djnh cua p h u o n g trinh t h i viec danh gia cua chung ta kho khan, d o i k h i la khong the danh gia v i mien ciia bien luc do rgng khong d a m bao cho viec danh gia. D o do rang buoc them dieu kien doi v o i nghiem cua p h u o n g trinh giup chiing ta thuan Igi trong danh gia t u do giai quye't duoc bai toan.

• ZBAlTAPLUYeNTAP _ ^ x W d n h i ^ n o . K l q r Bai 4.110: Giai cac bpt sau :

a) V x^ < 2 x - 1 c) V 3 x - 2 > 4x - 3

0 > S . - ^ C

' i . , . x , < 0 ^ b) N/X^ - x + 1 < x + 3 - " ^ ^ f '

a) Bpt

^) Bpt

2 x - l > 0

x - 3 > 0 <=>

x - 3 < ( 2 x - l ) 2

x^ - X +1 > 0 y x + 3 > 0

x 2- x + l < ( x + 3)^

d) V3x2+x-4>x + l

Hu&ng dan giai £y c> € 77% o (")

x > i ,,,an i\ *

x>3 o x> 3 ' 4x - 5 x + 4 > 0

• ằ x > — 7

351

c) Bpt<=> 4 x - 3 < 0 4 x - 3 > 0

3 x - 2 > 0 3 x - 2 > ( 4 x - 3 ) ' <=>

2 3

— ^ X < —

3 , 3 - < x < l , . .

4 )i.,;r; •(

d) Bpt<:>

x + l < 0 3 x ^ + x - 4 > 0

X + 1 >0 A ô u > { 3x^ + x - 4 > ( x + l)^

3

x> 1 + 74T.

Bai 4.111: Giai cac bat phuong trinh sau. 1 ^ / ' - f

a) (x^ - 3X)N/2X^ - 3 x - 2 > 0 - b) x^ + 3x +1 < (x + 3 ) N / X ^ Huang dan giai ' . ; a) Ta xet hai truong hop

T H 1: 2x^ - 3x - 2 = 0 ô x - 2,x = . Khi do BPT luon dung

fi' T H 2: B p t ô 2 x 2 - 3 - 2 > 0 x^ - 3 x > 0

x < - - V x > 2 1 ^ - 2 <=>x<— V x > 3 . x < 0 V x > 3 ^

' 1 - o c ;

V 2 u{2}u[3;+oo).

Vay nghi^m ciia Bpt da cho la: T =

b) Bat phuong trinh o x(x + 3) - (x + 3) 7 x 2+ l + l < 0 o ( x + 3 ) ( x - Vx^ +1) + (N/X^ + 1 )2 - X ^ <0

< ^ ( N / X2+ 1 - X J | N / X2+ 1- 3J < 0 (*)

Do Vx^ +1 - x > V x ^ - X = |x| - X > 0

= > ( * ) ằ V x^ + 1 <3<x>x2 < 8 o- 2 7 2 < X< 2N/ 2 .

V|y - 2 V 2 < x < 272 la nghi^m ciia bat phuong trinh da cho. ^ . Bai 4.112: Giai cac bpt sau : ^

a) V 2 x 2- 6 x + l - x + 2 > 0 c) Vx + 2 - Vx + 1 < N/X

b) Vx + 3 > V 2 x- 8 + N/7-X d) 2x^

( 3 - V 9 + 2x) Huang dan giai

:x + 21

x- 2 < 0

a) B p t< ^ V 2 x^ - 6 x . l > x- 2 o | ^ ^ , _ ^ ^ ^ ^ ^ ^

352

Cfi/ TNHHMTV DVVU K i u i H ^ Vift

hoac

' x - 2 > 0

• ( 2 x 2 - 6 x + l ) > ( x - 2 ) 2 ^ ' x < 2

x <

x > 3 + V7

hoac x > 2

x ^ - 2 x - 3 > 0 x <

x > 3 3 - V 7

b) DKXD:

x + 3 > 0 "

2 x - 8 > 0 < = > 4 < x < 7 7 - x > 0

Bpt ô x + 3> ( V 2 x- 8 + V 7 ^ ) o 3 > - l + 2 7 ( 2 xx2 - 8 ) ( 7 - x ) o 2 > 7 ( 2 x- 8 ) ( 7 - x ) o 4> -2x2 + 2 2 x- 5 6

X <5

X > 6

•4<x<5 ^ : o x ^ - l l x + 3 0 > 0 o

Doi chieu dieu kien ta nghi^m bpt la

6 < x < 7 c) D K X D :

X + 2 > 0 ' T • x + l> 0 <x>x>0

x> 0 ,„ ,

^pt O V X + 2<N/X + 1 + N ^ C J . X + 2 < 2X + 1 + 27(x + l ) x

. r i- x< o

< = > l - x < 2 J ( x + l ) x ô•< „ - - .

3 + 2^ ! f - < • j l - x > 0

hoac -! , o ( 1 - x ) <4x(x + l )

X < — - 3 + 2V3

•<x 1. / . - V'

^ 1 chieu dieu ki?n ta nghi^m bpt la x > ^ ' ^ ^ ^

^) D K X D : f9 + 2 x > 0

3 - 7 9 + 2 x ^ 0 2

X;tO

Bp 2x2(3 + ^/9T2^)

t o i < X + 21

4x2

Vv Jq<i !

oV9 + 2 x < 4 o x < - 2

9 7

— < X < —

2 2

X ^ 0 Doi chieu dieu i<ien ta nghiem bpt la

X ^ 0

- ^' ' v . ,

Bai 4.143: Giai cac bat phuong trinh sau: H ' \

b) \lx^ - 3x + 2 +\/x^ - 4 x + 3 >2\/x^ - 5x + 4^

Huang dan giai , V-3x^ + X + 4 + 2 -

a) — < ^

a) D K X D :

4 ! - 1 < x < -

x ^ O

Voi 0 < x < - : BPT ô 3

V_ 3 x 2^ x - f 4 + 2 ^ ^ ^ ^ _ 3^ 2^ ^ ^ 4 ^ 2 x - 2

"FfJ

2 x - 2 > 0 X> 1

_3x2+x + 4 < ( 2 x - 2 f [ 7 x 2 - 9 x> 0 9 o X > - 9 4

Suy ra nghiem cua ba't phuong trinh la -<x<- Voi - 1 < X < 0 : bpt luon diing

Doi chieu dieu ki?n ta c6 nghiem bpt la f x 2 - 3 x + 2 > 0

- l< x< 0 9 ^ 4

- < X < -

7 3

b) DKXD: x^ - 4 x + 3>0<:=>

x2 - 5 x + 4> 0

x> 4

x < l v S

0 < I

B p t o 7( x- l) ( x - 2 ) + V ( x- l) ( x- 3) > 2 7 ( x- l) ( x- 4 ) De tha'y x = 1 la nghiem ciia bpt.

+ Voi x < 1: Bpt c> 7(1 - x)(2 - x) + ^(1 - x)(3 - x) > 2^(1 - x)(4 - x) o N/ 2 -X + V 3- X > 2 V 4- X

Ta CO : >y2-x + V 3 - x < N / 4 -X + N/ 4 -X = 2 V 4- x Suy ra X < 1 bpt v6 nghiem .

+) Voi X > 4 : bpt o V x- 2 + N / X- 3 > 2>yx-4

T a c o : >yx - 2 + Vx - 3 > Vx - 4 + Vx - 4 = 2Vx - 4, Vx,x > 4 Suy ra : x > 4 bat pt luon dung . )

"x = l x > 4 Vay nghif m cua bpt la :

4.114: Giai cac bat phuong trinh sau:

g)V3x^+6x + 4< 2 - 2 x- x 2 b) 2x2 + 4x + sVs - 2x - x^

^) Vix^ + 5x + 7 - V 3 x 2+ 5 x + 2 > 1 d) Vx + 2 > / ^ + V x - 2 V x ^ > -

e) 5 V ^ . - f < 2 x . J - . 4 0 ^ - 2 . / ^ > 3 g ) x . ^ > 3 ^

2Vx 2x x + 1 V X 12

a) Dat

Huang dan giai rin/riJor'n at: t = V3x2 + 6x + 4 , t > 0= > x 2 +2x = ^—^ . K

Bat phuong trinh tro thanh

t < 2 ~ ô t 2 + 3 t - 1 0 < 0 o 0 < t< 2( t > 0 ) , . * ,

Ta CO V 3 x 2 +6x + 4 < 2 <ằ 3x^ + 6x + 4 < 4 o 3x2 + 6x < 0 o -2 < x < 0

Vay nghiem bpt la -2 < x < 0 .

b) DKXD: -3 < X < 1 ' ^ ' '

Dat: t = V 3 - 2 x- x 2, t > 0^ t 2= 3 - 2 x- x 2^ 2 x + x 2 = 3- t 2 <x Bat phuong trinh tro thanh i

2 ( 3 - t 2 ) + 3 t> l c^2t2 - 3 t- 5 < 0 c : > 0 < t < - ( d o t > 0 ) Ta CO V 3 - 2 x- x 2 < - ô .

2

-3 < X < 1

„ - 2 25 ô ' - 3 < x < l 3 - 2 x - x < —

xV

Vay nghiem bpt la -3 < x < 1.

Một phần của tài liệu phân loại và phương pháp giải đại số 10 (Trang 174 - 178)

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