\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{2}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : \(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{2}{x\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-1\right)}{x\left(x+1\right)}-\dfrac{x+1}{x\left(x+1\right)}=\dfrac{2}{x\left(x+1\right)}\)
`=> x(x-1) -(x+1)=2`
`<=>x^2 - x -x-1=2`
`<=> x^2 -2x-1+2=0`
`<=> x^2 -2x +1=0`
`<=> (x+1)^2=0`
`<=>x+1=0`
`<=>x=-1(ktm)`
Vậy pt vô nghiệm