\(\Leftrightarrow\frac{\left(x-1\right)^2}{x^2-1}-\frac{\left(x+1\right)^2}{x^2-1}=\frac{1-mx}{x^2-1}\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x+1\right)^2=1-mx\)
\(\Leftrightarrow-4x=1-mx\)
\(\Leftrightarrow x\left(m-4\right)=1\)
Với m=4 PT vô nghiệm
PT co nghiệm duy nhất : \(x=\frac{1}{m-4}\forall m\ne4\)
Ta có : \(\frac{x-1}{x+1}-\frac{x+1}{x-1}=\frac{1-mx}{x^2-1}\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{1-mx}{x^2-1}\)
\(\Leftrightarrow\frac{x^2-2x+1}{x^2-1}-\frac{x^2+2x+1}{x^2-1}=\frac{1-mx}{x^2-1}\)
\(\Leftrightarrow\frac{x^2-2x+1-x^2-2x-1}{x^2-1}-\frac{1-mx}{x^2-1}=0\)
\(\Leftrightarrow\frac{-4x}{x^2-1}-\frac{1-mx}{x^2-1}=0\)
\(\Leftrightarrow\frac{-4x-1+mx}{x^2-1}=0\)
\(\Leftrightarrow-4x-1+mx=0\)
\(\Leftrightarrow x\left(m-4\right)=1\)
