. x^, 4x +3 y, Ta thay phuong trinh c6 nghỉm khi va chi khi — va
a)V^+ ^32 x-y2 +3=0 b)
x/Zxy + Sx + S = 4 x y - 5 x - 3
Giai
a) Phuang trinh (1) tuong duong: Taco: ::' . . . i Taco: ::' . . . i VT = '82 2x + y j 6 + 80X' 6 + 80x ^ x 2 . . 1- 2x + y , > 9x + V2x+y 3 6 >9x+ . ^9x + '^/9(2x + y) 2x + y + 9 - o 3 x - 2 x 2 - x y + 6 y > 0 (*) 9 2x + y + 9
Lay (*) cQng voi PT(2) ta du^c:
- x 2 + 4 x y - 4 y 2+ 1 2 y- 6 x- 9 > 0 o - ( x - 2 y + 3)^ > 0 o x + 3 = 2 y . De dau bSng xay ra thi x = y = 3 .
Vay C O nghif m (x; y) = (3; 3 ) . b) Ta C O x + y 2 4 4 4 3 4 12 4 2 2 x + xy + y x + y T u do suy ra x2+y2 i — + x^ + xy + y^ x + y >x + y
Dau bang xay ra khi va chi khi x = y > 0 '
Thay x = y vao phuong trinh con lai ta c6: x^jlx^+5x + 3 = 4x2 - 5x - 3 De y rang x = 0 khong phai la nghi^m. Ta xet x > 0, chia phuong trinh cho
5 3 .
2 + - + — = 4 -'5 3 ^ . Dat t = x^ thi thu duQic:
phuang trinh: t 2+ t - 6 = 0 o t = 2 « 2 + - + 4 = 4 « 4 + - - 2 = 0<=>'' = 3 '2 + - + — > 0 ta C O X x2 X X ^ 3 5 x2 X
W^y h^ phuong trinh c6 nghi^m duy nhat (x; y) = (3; 3)
V i d^ 4: Giki cac h ? phuang trinh sau
a) V^ + ^ 3 2 - x- y 2+ 3 = 0 b) b) ^ + V 3 2 - x + 6 y - 2 4 = 0 ^ + ( x - y ) ( V x y - 2 ) + %/x = y + J 7 (x + 1) y + yfxy + xil-x) =4 (1) (2) 1- A J. Giai 0 < x < 32 [ y < 4
Cpng hai phuang trinh ve theo ve ta c6: a) Dieu ki^n:
>^ + V32 - X + ^ +1/32 - x = y2 - 6y + 21 (*) Ta c6: y2 - 6y + 21 = (y - 3)^ +12 > 12 .
Mat khac theo bat dang thuc Bunhiacopxki ta c6:
> y x + V 3 2 - X < ^ ( l + l)( x + 3 2 - x ) =8
+ t/ 3 2-x < ^(l + l)(V^ + V 3 2- x ) - 4
Vay Vx + V 3 2 - X + ^ + t/32-x < 12. Tir do suy ra h? c6 nghi^m khi va chi
khi x,y phai thoa man:
= V32 - X ^ = t / 3 2 - x o . y - 3 = 0
x = 16 y = 3
Vay h^ phuong trinh c6 1 nghi^m duy nhat (x;y) = (16;3)
Ngoai each dimg bat dang thuc ta cung c6 the diing ham só dói voi f(x) = ^ + N / 3 2 - X + ^ + ^ 3 2 - X tren [O;32]
That vay taco: ^'
^.^ ^ 1 1 ^ 1 1 ^4^:^-4^
2N / 3 2 ^ ^ 4 V ? 44/(32-x)^ 2Vx(32-x) ^ 4 ^ ^ ( 3 2 ^ Nhan xet: V 3 2 - x - Vx luon cung dau voi V 3 2 - x - \/x
Suy ra f (x) = 0 <=> x = 16
De y rang: f(0) = f(32) = 732 + t/32;f(16) = 12 = ^ N/32 + ^32 < f(x) < 12 9fl .
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b) D i e u k i f n :' ^ |xy + ( x - y ) [ V x y - 2 ) > 0 / „ \ .11—
Chuyen ve va b i n h p h u a n g 6 p h u a n g trinh t h u nhát cua h? ta t h u dxxgc:
xy + ( x - y ) ( V ^ - 2 ) = (y + Vy-Vx)^ \< » ( x - y ) ( y + ^ - 2 ) + ( V x - ^ ) ( 2 y + 7y-NĂ) = 0 (3) < » ( x - y ) ( y + ^ - 2 ) + ( V x - ^ ) ( 2 y + 7y-NĂ) = 0 (3) T u p h u a n g trinh (1) c u a h ? t a c 6 2y + J y - V x = y + ^ + ( x - y ) ( ^ - 2 ) ^ 0 . T u p h u a n g trinh (2) ta c6: (x + l ) ( y + 7xy) = x ^ - x + 4 = (x + 2 ) ( x - l ) 2 + 2 ( x + l ) > 2 ( x + l ) = > y + 7 ^ s 2 Ket hgrp v a i (3) ta suy ra x = y Thay vac p h u a n g trinh (2) ta c6:
(x + l ) [ 2 x + x ( l - x ) ] = 4 o x ^ - 2 x 2 - 3 x + 4 = 0 o x = l
Ket lủin: c6 nghỉm d u y nhát x = y = 1 f
N h ^ n xet: Vỉc n h i n ra du(?c quan h? x = y la chia khoa de giai quyét bai toan. Day la ky nang dac bỉt quan trpng k h i giai h? bang p h u a n g phap danh gia ciing n h u chung m i n h bat dang thuc.
V i d v 5: G i a i cac h$ p h u a n g t r i n h sau a) • e" + (x^ - y ) l n ( x + y + 2014) = ê^ (1) a) • 9 ' < + 4 x _ 3 x _ 2 ' ' ^ 3 4 y _ 2 6 x (2) b) < 3 " + 3 ^ + 3 ^ = 5 (2) ' Giii a) T u p h u a n g trinh (1) suy ra y > 0 .
Ta Viet l^i p h u o n g trinh (2) thanh: 2''(2'' - 1 ) + 3''(3'' - 1 ) = 34y - 26x 2 ' ' < 1
+ N e u X < 0:
3" < 1
suy ra 2 ' * ( 2 ' ' - l ) + 3 ' ' ( 3 ' ' - l ) < O n h u n g 3 4 y - 2 6 x > 0
Vay h^ v6 nghi^m. N h u v^y h? c6 nghi^m k h i x , y > 0 Ta viet lai p h u a n g trinh (1) cua h? thanh:
e ' ' - e ^ + ( x 2 - y ) l n ( x + y + 2014) = 0 (*)
V i h a m só f(x) = e" dong bien tren R nen e" - e ^ cung dáu v o i x - ^
174
M ^ t khac l n ( x + y + 2014) > 0 nen t u (*) ta suy ra x = ^ <=> x = y^ thay vao phuang trinh (2) ta c6:
9'< + 4''=3''+2''+34x2-26x « 9" + 4" - 3" - 2" - 34x2 + 26x = 0 Xet ham so f(x) = 9=^ + 4" - 3" - 2" - 34x2 ^ 26x tren [0;+oo) ta c6
f (x) = 9" + 4" - 3" - 2" - 68x + 26;
r(x) = 9 ' ' + 4' ' - 3' ' - 2 ' ' - 6 8 ;
r( x ) = 9 ' ' + 4' ' - 3' ' - 2 ' ' > 0
L9P bang bien thien cua ham so ta suy ra f(x) = 0 c6 toi da 3 n g h i f m
M$t khac ta thay f(0) = f ( l ) = f(2) = 0 suy ra p h u a n g trinh c6 d i i n g 3 nghi^m x = 0;x = l ; x = 2
V$y h? p h u a n g trinh CO 3c|pnghỉm: ( x ; y ) = ( 0 ; 0 ) , ( l ; l ) , ( 2 ; 4 ) b) T u p h u a n g trinh (1) cua h? ta suy ra x , y , z e [ 0 ; l "
Truoc hét ta chung m i n h : 3' > 21^ + 1 (*) v a i m p i t e [ 0 ; l " T h a t v a y x e t h a m s o f(t) = 3' - 2 t 2 + l tren [ 0 ; l "
Ta CO f'(t) = 3' In 3 - 4t; f ' ( t ) = 3' In^ 3- 4 ; f "(t) = 3' In^ 3 > 0 N h u vay h a m so f(t) = 0 c6 toi da 3 nghi^m.
Mat khac f(0) = £(1) = f(2) = 0 nen f(t) = 0 c6 toi da 3 nghỉm la t = 0; t = 1; t = 2
De y ta thay h a m so f(t) lien tyc tren [ 0 ; l ] va f(0) = f ( l ) = 0 nen f(t) khong doi dáu tren f O ; !
Do f ^ 1 ^ = ^ / 3 - | > 0 = > f ( t ) S 0 V t € [ 0 ; l to/.
A p d y n g (*) ta suy ra 3" + 3^ + 3^ > 2(x2 + y^ + z2 J ^ 1 = 5
Dáu bang xay ra k h i va chi k h i ] (adsbygoogle = window.adsbygoogle || []).push({});
[ x , y , z e | 0 ; l