Khi giải phương trình \(\dfrac{1}{x}+2=\left(\dfrac{1}{x}+2\right)\left(x^2+1\right)\) , ta được nghiệm là
\(x=-\dfrac{1}{2}\).\(x=2\).\(x=\dfrac{1}{2}\).\(x=0\) hoặc \(x=-\dfrac{1}{2}\).Hướng dẫn giải:Đk: \(x\ne0\)
\(\dfrac{1}{x}+2=\left(\dfrac{1}{x}+2\right)\left(x^2+1\right)\)
\(\Leftrightarrow\left(\dfrac{1}{x}+2\right)\left(x^2+1-1\right)=0\)
\(\Leftrightarrow x^2\left(\dfrac{1}{x}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\\dfrac{1}{x}+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
