$\text{Đáp án+Giải thích các bước giải:}$
$\text{ĐKXĐ: x$\neq$-y , y$\neq$1 ,x$\neq$-1}$
$\text{Có P = $\frac{x²}{(x+y)(1-y)}$ - $\frac{y²}{(x+y)(1+x)}$ - $\frac{x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{1}{x+y}$( $\frac{x²}{1-y}$ - $\frac{y²}{1+x}$ )- $\frac{x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{1}{x+y}$.$\frac{x³+x²+y³-y²}{(1+x)(1-y)}$ -$\frac{x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{1}{x+y}$.$\frac{(x-y)(x+y)+(x+y)(x²-xy+y²)}{(1+x)(1-y)}$ - $\frac{x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{1}{x+y}$.$\frac{(x+y)(x²+x+y²-xy-y}{(1+x)(1-y)}$ - $\frac{x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{x²+x+y²-xy-y}{(1+x)(1-y)}$ - $\frac{x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{x²+x+y²-xy-y-x²y²}{(1+x)(1-y)}$}$
$\text{= $\frac{(x-y)+x(x-y)+y²(1-x²)}{(1+x)(1-y)}$}$
$\text{= $\frac{(x-y)(x+1)+y²(1-x)(1+x)}{(1+x)(1-y)}$}$
$\text{= $\frac{(x+1)(x-y-xy²+y²)}{(1+x)(1-y)}$}$
$\text{= $\frac{(x+1)(1-y)(x+xy-y)}{(1+x)(1-y)}$}$
$\text{= x-y+xy}$
$\text{Vậy P=x-y+xy}$
