\(\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(x-\dfrac{1}{x}\right)\) (1)
ĐK: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\\x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne0\end{matrix}\right.\) \(\Leftrightarrow x\ne\pm1;x\ne0\)
\(\left(1\right)=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(\dfrac{x^2}{x}-\dfrac{1}{x}\right)\)
\(=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{x^2-1}{x}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{x}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{x}\cdot\dfrac{1}{x-1}+\dfrac{\left(x+1\right)\left(x-1\right)}{x}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{x+1}{x}+\dfrac{x-1}{x}\)
\(=\dfrac{x+1+x-1}{x}\)
\(=\dfrac{2x}{x}\)
\(=2\)
