Đk tự đặt nhé!Pt $\Leftrightarrow \sqrt{\frac{1-x}{x}}-1=\frac{2x+x^2}{1+x^2}-1$
$\Leftrightarrow \frac{\frac{1-x}{x}-1}{\sqrt{\frac{1-x}{x}}+1}=\frac{2x-1}{1+x^2}$
$\Leftrightarrow \frac{2x-1}{1+x^2}+\frac{2x-1}{x(\sqrt{\frac{1-x}{x}}+1)}=0$
$\Leftrightarrow (2x-1)(\frac{1}{1+x^2}+\frac{1}{x(\sqrt{\frac{1-x}{x}}+1)})=0$