a/ Để m có nghĩa thì:
\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ne0\\2\left(1-x^2\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ne0\\x^2\ne1\end{matrix}\right.\)\(\Leftrightarrow x\ne\pm1\)
b/ \(M=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}=\dfrac{x}{2\left(x-1\right)}+\dfrac{-\left(x^2+1\right)}{2x^2-2}\)
\(=\dfrac{x}{2\left(x-1\right)}+\dfrac{-x^2-1}{2\left(x^2-1\right)}=\dfrac{x}{2\left(x-1\right)}+\dfrac{-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\)
c/ M = \(-\dfrac{1}{2}\) \(\Leftrightarrow\dfrac{1}{2\left(x+1\right)}=-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{-1}{2}\cdot2\left(x+1\right)=1\Leftrightarrow-x-1=1\Leftrightarrow-x=2\Leftrightarrow x=-2\)
