\(\Leftrightarrow\sqrt{\frac{x+2}{x}}=-2x-4+\frac{3}{x}\)
- Với \(x>0\), đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\\\sqrt{\frac{1}{x}}=b\end{matrix}\right.\)
\(\Rightarrow ab=-2a^2+3b^2\)
\(\Leftrightarrow2a^2+ab-3b^2=0\Leftrightarrow\left(a-b\right)\left(2a+3b\right)=0\)
\(\Leftrightarrow a=b\Leftrightarrow x+2=\frac{1}{x}\Leftrightarrow x^2+2x-1=0\)
- Với \(x\le-2\), đặt \(\left\{{}\begin{matrix}\sqrt{-x-2}=a\\\sqrt{-\frac{1}{x}}=b\end{matrix}\right.\)
\(\Rightarrow ab=2a^2-3b^2\Leftrightarrow2a^2-ab-3b^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(2a-3b\right)=0\Leftrightarrow2a=3b\)
\(\Leftrightarrow4\left(-x-2\right)=-\frac{9}{x}\Leftrightarrow...\)
