Đáp án:

\(\begin{array}{l}
1,\\
{A_1} = \dfrac{{4{x^2} + x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
2,\\
{A_2} = \dfrac{{ - 10{a^2} + 4ab + 3{b^2}}}{{12ab}}\\
3,\\
{A_3} = \dfrac{1}{{x + 1}}\\
4,\\
{A_4} = \dfrac{{ - 1}}{x}\\
5,\\
{A_5} = x + 2\\
6,\\
{A_6} = \dfrac{1}{{x + 1}}\\
7,\\
{A_7} = \dfrac{{8x + 16}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
8,\\
{A_8} = 0
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 2 \ne 0\\
2 - 2{x^2} \ne 0
\end{array} \right. \Leftrightarrow x \ne  \pm 1\\
{A_1} = \dfrac{x}{{2x - 2}} + \dfrac{{ - {x^2} - 1}}{{2 - 2{x^2}}} + 1\\
 = \dfrac{x}{{2x - 2}} + \dfrac{{{x^2} + 1}}{{2{x^2} - 2}} + 1\\
 = \dfrac{x}{{2\left( {x - 1} \right)}} + \dfrac{{{x^2} + 1}}{{2\left( {{x^2} - 1} \right)}} + 1\\
 = \dfrac{x}{{2\left( {x - 1} \right)}} + \dfrac{{{x^2} + 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} + 1\\
 = \dfrac{{x.\left( {x + 1} \right) + \left( {{x^2} + 1} \right) + 2.\left( {x - 1} \right)\left( {x + 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + x + {x^2} + 1 + 2.\left( {{x^2} - 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + x + {x^2} + 1 + 2{x^2} - 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{4{x^2} + x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
2,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
3ab \ne 0\\
6b \ne 0\\
12a \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ne 0\\
b \ne 0
\end{array} \right.\\
{A_2} = \dfrac{{ - {a^2} + {b^2}}}{{3ab}} - \dfrac{{3a - 2b}}{{6b}} - \dfrac{b}{{12a}}\\
 = \dfrac{{4.\left( { - {a^2} + {b^2}} \right) - 2a\left( {3a - 2b} \right) - b.b}}{{12ab}}\\
 = \dfrac{{\left( { - 4{a^2} + 4{b^2}} \right) - \left( {6{a^2} - 4ab} \right) - {b^2}}}{{12ab}}\\
 = \dfrac{{ - 4{a^2} + 4{b^2} - 6{a^2} + 4ab - {b^2}}}{{12ab}}\\
 = \dfrac{{ - 10{a^2} + 4ab + 3{b^2}}}{{12ab}}\\
3,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 2 \ne 0\\
2{x^2} - 2 \ne 0
\end{array} \right. \Leftrightarrow x \ne  \pm 1\\
{A_3} = \dfrac{{x + 1}}{{2x - 2}} - \dfrac{{{x^2} + 3}}{{2{x^2} - 2}}\\
 = \dfrac{{x + 1}}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 3}}{{2\left( {{x^2} - 1} \right)}}\\
 = \dfrac{{x + 1}}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2} - \left( {{x^2} + 3} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{\left( {{x^2} + 2x + 1} \right) - \left( {{x^2} + 3} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1 - {x^2} - 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2x - 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2\left( {x - 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{1}{{x + 1}}\\
4,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2{x^2} + 6x \ne 0\\
2x + 6 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne  - 3
\end{array} \right.\\
{A_4} = \dfrac{{x - 6}}{{2{x^2} + 6x}} - \dfrac{3}{{2x + 6}}\\
 = \dfrac{{x - 6}}{{x\left( {2x + 6} \right)}} - \dfrac{3}{{2x + 6}}\\
 = \dfrac{{\left( {x - 6} \right) - 3.x}}{{x.\left( {2x + 6} \right)}}\\
 = \dfrac{{x - 6 - 3x}}{{x\left( {2x + 6} \right)}}\\
 = \dfrac{{ - 2x - 6}}{{x\left( {2x + 6} \right)}}\\
 = \dfrac{{ - \left( {2x + 6} \right)}}{{x\left( {2x + 6} \right)}}\\
 = \dfrac{{ - 1}}{x}\\
5,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x - 6 \ne 0\\
x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 6\\
x \ne  - 1
\end{array} \right.\\
{A_5} = \dfrac{{2{x^2} - 12x}}{{x - 6}} - \dfrac{{1 - {x^2}}}{{x + 1}} + 3 - 2x\\
 = \dfrac{{2x.x - 2x.6}}{{x - 6}} - \dfrac{{{1^2} - {x^2}}}{{x + 1}} + 3 - 2x\\
 = \dfrac{{2x.\left( {x - 6} \right)}}{{x - 6}} - \dfrac{{\left( {1 - x} \right)\left( {1 + x} \right)}}{{x + 1}} + 3 - 2x\\
 = 2x - \left( {1 - x} \right) + 3 - 2x\\
 = 2x - 1 + x + 3 - 2x\\
 = x + 2\\
6,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 2 \ne 0\\
2 - 2{x^2} \ne 0
\end{array} \right. \Leftrightarrow x \ne  \pm 1\\
{A_6} = \dfrac{{x + 1}}{{2x - 2}} + \dfrac{{{x^2} + 3}}{{2 - 2{x^2}}}\\
 = \dfrac{{x + 1}}{{2x - 2}} - \dfrac{{{x^2} + 3}}{{2{x^2} - 2}}\\
 = \dfrac{{x + 1}}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 3}}{{2\left( {{x^2} - 1} \right)}}\\
 = \dfrac{{x + 1}}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2} - \left( {{x^2} + 3} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{\left( {{x^2} + 2x + 1} \right) - \left( {{x^2} + 3} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{{x^2} + 2x + 1 - {x^2} - 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2x - 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{{2\left( {x - 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
 = \dfrac{1}{{x + 1}}\\
7,\\
DKXD:\,\,\left\{ \begin{array}{l}
x - 5 \ne 0\\
25 - {x^2} \ne 0\\
x + 5 \ne 0
\end{array} \right. \Leftrightarrow x \ne  \pm 5\\
{A_7} = \dfrac{6}{{x - 5}} - \dfrac{{6 - 2x}}{{25 - {x^2}}} + \dfrac{4}{{x + 5}}\\
 = \dfrac{6}{{x - 5}} - \dfrac{{2x - 6}}{{{x^2} - 25}} + \dfrac{4}{{x + 5}}\\
 = \dfrac{6}{{x - 5}} - \dfrac{{2x - 6}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} + \dfrac{4}{{x + 5}}\\
 = \dfrac{{6.\left( {x + 5} \right) - \left( {2x - 6} \right) + 4.\left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 = \dfrac{{\left( {6x + 30} \right) - \left( {2x - 6} \right) + \left( {4x - 20} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 = \dfrac{{6x + 30 - 2x + 6 + 4x - 20}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
 = \dfrac{{8x + 16}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
8,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
{x^2} - 4 \ne 0\\
2x + 4 \ne 0\\
4 - 2x \ne 0
\end{array} \right. \Leftrightarrow x \ne  \pm 2\\
{A_8} = \dfrac{{4x}}{{{x^2} - 4}} + \dfrac{{x - 2}}{{2x + 4}} + \dfrac{{x + 2}}{{4 - 2x}}\\
 = \dfrac{{4x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{{x - 2}}{{2x + 4}} - \dfrac{{x + 2}}{{2x - 4}}\\
 = \dfrac{{4x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{{x - 2}}{{2\left( {x + 2} \right)}} - \dfrac{{x + 2}}{{2\left( {x - 2} \right)}}\\
 = \dfrac{{2.4x + {{\left( {x - 2} \right)}^2} - {{\left( {x + 2} \right)}^2}}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{8x + \left( {{x^2} - 4x + 4} \right) - \left( {{x^2} + 4x + 4} \right)}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{8x + {x^2} - 4x + 4 - {x^2} - 4x - 4}}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = \dfrac{0}{{2\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = 0
\end{array}\)