$\frac{2}{x^2+2x+1}-\frac{5}{x^2-2x+1}=\frac{3}{1-x^2}$
$(ĐK: x\ne \pm1)$
$⇔\frac{2}{(x+1)^2}-\frac{5}{(x-1)^2}=-\frac{3}{(x-1)(x+1)}$
$⇔\frac{2(x-1)^2}{(x+1)^2(x-1)^2}-\frac{5(x+1)^2}{(x-1)^2(x+1)^2}+\frac{3(x-1)(x+1)}{(x-1)^2(x+1)^2}=0$
$⇔2(x-1)^2-5(x+1)^2+3(x^2-1)=0$
$⇔2(x^2-2x+1)-5(x^2+2x+1)+3x^2-3=0$
$⇔2x^2-4x+2-5x^2-10x-5+3x^2-3=0$
$⇔-14x-6=0$
$⇔x=-\frac{3}{7}$ $\texttt{(thỏa mãn)}$
