Đáp án:
$\begin{array}{l}
7)\\
a)Đkxđ:\left\{ \begin{array}{l}
2x - 2 \ne 0\\
2 - 2{x^2} \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne \pm 1
\end{array} \right. \Rightarrow x \ne \pm 1\\
b)C = \frac{x}{{2x - 2}} + \frac{{{x^2} + 1}}{{2 - 2{x^2}}}\\
= \frac{x}{{2\left( {x - 1} \right)}} - \frac{{{x^2} + 1}}{{2\left( {{x^2} - 1} \right)}}\\
= \frac{{x\left( {x + 1} \right) - \left( {{x^2} + 1} \right)}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \frac{{x - 1}}{{2\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \frac{1}{{2\left( {x + 1} \right)}}\\
c)Đkxđ:x \ne \pm 1\\
C = \frac{{ - 1}}{2}\\
\Rightarrow \frac{1}{{2\left( {x + 1} \right)}} = \frac{{ - 1}}{2}\\
\Rightarrow x + 1 = - 1\\
\Rightarrow x = - 2\left( {tmdk} \right)
\end{array}$
Vậy x=-2 thì thỏa mãn đề bài
