Question

tìm x,y thuộc z thỏa mãn:

a)2y-3 = \(\dfrac{2x+1}{x-2}\)

b)(y-1).(x² + x) = 2x

Answer 1

a) 2y-3 =\(\dfrac{2x+1}{x-2}\)

Vì x,y thuộc z nên: 2x+1 \(⋮\) x-2

=> 2(x-2)+5 \(⋮\) x-2

Mà 2(x-2) \(⋮\) x-2 => 5\(⋮\) x-2 => x-2\(\in\) Ư(5)

=>x-2\(\in\)\(\left\{1;-1;5;-5\right\}\)

=> x \(\in\)\(\left\{3;1;7;-3\right\}\)

Thay x vào ,ta có :

	x 3 1 7 -3 2y-3 7 -3 3 1 y 5 0 3 2

Answer 2

b) (y-1)(x²+x) =2x

=>y-1= \(\dfrac{2x}{x^2+x}=\dfrac{2x}{\left(x+1\right)x}\)

=> y-1 =\(\dfrac{2}{x+1}\)

=>(y-1)(x+1)=2

Mà 2=1.2=-1.(-2)

Ta có:

y-1	1	2	-1	-2
x+1	2	1	-2	-1
y	2	3	0	-1
x	1	0	-3	-2

Vậy các cặp (y;x) là: (2;1),(3;0),(0:-3),(-1;-2)