Đáp án:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ne 1\\
x \ne - 1
\end{array} \right.\\
b,\\
B = 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 2 \ne 0\\
{x^2} - 1 \ne 0\\
2x + 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2\left( {x - 1} \right) \ne 0\\
\left( {x - 1} \right)\left( {x + 1} \right) \ne 0\\
2\left( {x + 1} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - 1 \ne 0\\
x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne - 1
\end{array} \right.\\
b,\\
B = \left( {\dfrac{{x + 1}}{{2x - 2}} + \dfrac{3}{{{x^2} - 1}} - \dfrac{{x + 3}}{{2x + 2}}} \right).\dfrac{{4{x^2} - 4}}{5}\\
= \left( {\dfrac{{x + 1}}{{2\left( {x - 1} \right)}} + \dfrac{3}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{x + 3}}{{2\left( {x + 1} \right)}}} \right).\dfrac{{4.\left( {{x^2} - 1} \right)}}{5}\\
= \dfrac{{{{\left( {x + 1} \right)}^2} + 3.2 - \left( {x + 3} \right)\left( {x - 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{4.\left( {x - 1} \right)\left( {x + 1} \right)}}{5}\\
= \dfrac{{\left( {{x^2} + 2x + 1} \right) + 6 - \left( {{x^2} + 2x - 3} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{4.\left( {x - 1} \right)\left( {x + 1} \right)}}{5}\\
= \dfrac{{{x^2} + 2x + 1 + 6 - {x^2} - 2x + 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{4.\left( {x - 1} \right)\left( {x + 1} \right)}}{5}\\
= \dfrac{{10}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{5}\\
= \dfrac{{10.4}}{{2.5}}\\
= 4
\end{array}\)
