Đáp án:
Điều kiện xác định: x$\neq$±1
$\frac{x^2+4x+4}{2x+4}$ - $\frac{4x^2-8x+4}{x^2-1}$ = $\frac{(x+2)^2}{2(x+2)}$ - $\frac{4(x^2-2x+1)}{(x-1)(x+1)}$ = $\frac{x+2}{2}$ - $\frac{4(x-1)^2}{(x-1)(x+1)}$ = $\frac{x+2}{2}$ - $\frac{4(x-1)}{x+1}$ = $\frac{(x+2)(x+1)}{2(x+1)}$ - $\frac{4.2(x-1)}{2(x+1)}$ = $\frac{x^2+x+2x+2}{2(x+1)}$ - $\frac{8x-8}{2(x+1)}$ = $\frac{x^2+x+2x+2-8x+8}{2(x+1)}$ = $\frac{x^2-5x+10}{2(x+1)}$
