ĐK: \(x\ne0\)
Đặt \(x+\frac{1}{x}=a\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}+2=a^2\)
\(\Leftrightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(pt\Leftrightarrow2\left(a^2-2\right)-3a+2=0\)
\(\Leftrightarrow2a^2-4+3a+2=0\)
\(\Leftrightarrow2a^2+3a-2=0\)
\(\Leftrightarrow2a^2+4a-a-2=0\)
\(\Leftrightarrow2a\left(a+2\right)-\left(a+2\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(2a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-2\\a=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-2\\x+\frac{1}{x}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\varnothing\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=-1\)
