a) Để M có nghĩa thì \(x\ne1;x\ne-1\)
b) \(M=\dfrac{x}{2x-2}+\dfrac{x^2-1}{2-2x^2}\)
\(=\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{2x-2}\)
c) Để \(M=\dfrac{1}{2}\) thì \(\dfrac{1}{2x-2}=\dfrac{1}{2}\)
\(\Leftrightarrow2x-2=2\)
\(\Leftrightarrow x=2\)