a, Để C có nghĩa.
$⇔2x-2\neq0⇔x\neq1$
$2-2x^2\neq0⇔x\neq±1$
$⇒x\neq±1$.
b, $C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}$ ĐKXĐ: $x\neq±1$
$C=\dfrac{x(x+1)-x^2-1}{2(x-1)(x+1)}$
$C=\dfrac{x^2+x-x^2-1}{2(x-1)(x+1)}$
$C=\dfrac{x-1}{2(x-1)(x+1)}$
$C=\dfrac{1}{2(x+1)}$
Vậy $C=\dfrac{1}{2(x+1)}$ với $x\neq±1$.
c, Để $C=\dfrac{-1}{2}$
$⇔\dfrac{1}{2(x+1)}=\dfrac{-1}{2}$
$⇔\dfrac{1}{2(x+1)}-\dfrac{-x-1}{2(x+1)}=0$
$⇔\dfrac{1+x+1}{2(x+1)}=0$
$⇒x+2=0$
$⇔x=-2(tm)$
Vậy để $C=\dfrac{-1}{2}$ thì $x=-2$.