ĐKXĐ: ...
Đặt \(\sqrt{\frac{x+1}{x}}=t>0\)
\(\Rightarrow\frac{1}{t^2}-2t=3\)
\(\Leftrightarrow2t^3+3t^2-1=0\)
\(\Leftrightarrow\left(t+1\right)^2\left(2t-1\right)=0\)
\(\Leftrightarrow t=\frac{1}{2}\Leftrightarrow\frac{x+1}{x}=\frac{1}{4}\)
\(\Leftrightarrow...\)