Cho biểu thức: \(C = \frac{x}{{2x - 2}} + \frac{{{x^2} + 1}}{{2 - 2{x^2}}}\).

a) Tìm ĐKXĐ.

b) Rút gọn C.

c) Tìm x để \(C = \frac{{ - 1}}{2}\).

a) ĐKXĐ: \(\left\{ \begin{array}{l}2x - 2 \ne 0\\2 - 2{x^2} \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 1\\{x^2} \ne 1\end{array} \right. \Leftrightarrow x \Leftrightarrow \pm 1\)

b) \(C = \frac{x}{{2x - 2}} + \frac{{{x^2} + 1}}{{2 - 2{x^2}}}\)

\(C = \frac{x}{{2\left( {x - 1} \right)}} + \frac{{{x^2} + 1}}{{2\left( {1 - {x^2}} \right)}}\)

\(C = \frac{{x\left( {x + 1} \right) - {x^2} - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\)

\(C = \frac{{x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\)

\(C = \frac{1}{{2\left( {x + 1} \right)}}\)

c) Để \(C = \frac{{ - 1}}{2}\) thì \(\frac{1}{{2\left( {x + 1} \right)}} = \frac{{ - 1}}{2}\)

Suy ra: x + 1 = – 1

Hay x = –1 – 1 = –2

Vậy x = –2 thì \(C = \frac{{ - 1}}{2}\)