ĐKXĐ:\(x\ne0\)
\(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\\ \Leftrightarrow\dfrac{x^3}{x^2}+\dfrac{x}{x^2}=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\ \Leftrightarrow\dfrac{x^4}{x^2}+\dfrac{1}{x^2}-\dfrac{x^3}{x^2}-\dfrac{x}{x^2}=0\\ \Leftrightarrow\dfrac{x^4-x^3-x+1}{x^2}=0\\ \Leftrightarrow x^3\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\\left(x+\dfrac{1}{4}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\end{matrix}\right.\)
ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow x+\dfrac{1}{x}=\left(x^2+\dfrac{1}{x^2}+2\right)-2\)
\(\Leftrightarrow x+\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^2-2\)
Đặt \(x+\dfrac{1}{x}=t\Rightarrow t=t^2-2\)
\(\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-2x+1=0\end{matrix}\right.\)
\(\Rightarrow x=1\)
