Question

Giai PT

\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{-1}{x\left(x+1\right)}\)

Answer 1

\(\Leftrightarrow x^2-x-x-1=-1\)

=>x(x-2)=0

=>x=2

Answer 2

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)

\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow\dfrac{x\left(x-1\right)}{x\left(x+1\right)}-\dfrac{\left(x+1\right)}{x\left(x+1\right)}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2-x-x-1}{x\left(x+1\right)}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow x^2-2x-1=-1\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)