Giải thích các bước giải:
Ta có:
$A=\dfrac{x^2}{(x+y)(1-y)}-\dfrac{y^2}{(x+y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{x^2(1+x)-y^2(1-y)}{(x+y)(1-y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{x^3+y^3+x^2-y^2}{(x+y)(1-y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{(x+y)(x^2-xy+y^2)+(x-y)(x+y)}{(x+y)(1-y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{(x+y)(x^2-xy+y^2+x-y)}{(x+y)(1-y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{x^2-xy+y^2+x-y}{(1-y)(1+x)}-\dfrac{x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{x^2-xy+y^2+x-y-x^2y^2}{(1+x)(1-y)}$
$\to A=\dfrac{x^2+y^2-x^2y^2+x-y-xy}{(1+x)(1-y)}$
$\to A=\dfrac{x^2+y^2-x^2y^2-1+x-y-xy+1}{(1+x)(1-y)}$
$\to A=\dfrac{(x^2-1)(1-y^2)+(1+x)(1-y)}{(1+x)(1-y)}$
$\to A=\dfrac{(x-1)(x+1)(1-y)(1+y)+(1+x)(1-y)}{(1+x)(1-y)}$
$\to A=(x-1)(1+y)+1$
Để $A=2$
$\to (x-1)(1+y)+1=2$
$\to (x-1)(1+y)=1$
$\to (x-1,1+y)$ là cặp ước của $1$ vì $x,y\in Z$
$\to (x-1,1+y)\in\{(1,1), (-1,-1)\}$
$\to (x,y)\in\{(2, 0), (0,-2)\}$
