Bài giải:

Điều kiện:$x\ne 1;x\ne -1$

$\frac{2x+1}{{{x}^{2}}-2x+1}-\frac{2x+3}{{{x}^{2}}-1}=0$

$\Leftrightarrow \frac{2x+1}{{{\left( x-1 \right)}^{2}}}-\frac{2x+3}{\left( x-1 \right)\left( x+1 \right)}=0$

$\Leftrightarrow \frac{\left( 2x+1 \right)\left( x+1 \right)-\left( 2x+3 \right)\left( x-1 \right)}{{{\left( x-1 \right)}^{2}}\left( x+1 \right)}=0$

$\Leftrightarrow \frac{2{{x}^{2}}+2x+x+1-2{{x}^{2}}+2x-3x+3}{\left( x+1 \right){{\left( x-1 \right)}^{2}}}=0$

$\Leftrightarrow \frac{2x+4}{\left( x+1 \right){{\left( x-1 \right)}^{2}}}=0$

$\Leftrightarrow 2x+4=0$

$\Leftrightarrow x=-2$

Trả lời: x=-2