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

a) Rút gọn biểu thức A.

b) Tìm giá trị của x để A > –1.

a) Điều kiện xác định: \(\left\{ \begin{array}{l}2x - 2 \ne 0\\2 - 2{x^2} \ne 0\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}x \ne 1\\x \ne \pm 1\end{array} \right.\)

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

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

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

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

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

A = \[\frac{{x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{2\left( {x + 1} \right)}}\] (x ≠ ±1)

b) Để A > –1 tức \[\frac{1}{{2\left( {x + 1} \right)}} > - 1\]

⇔ \[\frac{1}{{2\left( {x + 1} \right)}} + 1 > 0\]

⇔ \[\frac{{1 + 2x + 2}}{{2\left( {x + 1} \right)}} > 0\]

⇔ \[\frac{{1 + 2x + 2}}{{2\left( {x + 1} \right)}} > 0\]

⇔ \[\frac{{2x + 3}}{{2\left( {x + 1} \right)}} > 0\]

⇔ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}2x + 3 > 0\\2x + 2 > 0\end{array} \right.\\\left\{ \begin{array}{l}2x + 3 < 0\\2x + 2 < 0\end{array} \right.\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x > - 1\\x < \frac{{ - 3}}{2}\end{array} \right.\)

Vậy để A > –1 thì x ∈ (–∞; \(\frac{{ - 3}}{2}\)) ∪ (–1; +∞)\{1}