Đáp án:

\(\begin{array}{l}
1,\\
a,\\
A = \dfrac{{ - 1}}{{1 + a + {a^2}}}\\
a = \sqrt 2  \Rightarrow A = \dfrac{{\sqrt 2  - 3}}{7}\\
b,\\
B = \left| {\sqrt {{a^2} - 1}  + 1} \right| - \left| {\sqrt {{a^2} - 1}  - 1} \right|\\
a = \sqrt 5  \Rightarrow B = 2\\
2,\\
a,\\
M = \dfrac{{\sqrt a  - 1}}{{\sqrt a }}\\
b,\\
M < 1,\,\,\,\forall a > 0,a \ne 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right.\\
A = \dfrac{1}{{2\left( {1 + \sqrt a } \right)}} + \dfrac{1}{{2\left( {1 - \sqrt a } \right)}} - \dfrac{{{a^2} + 2}}{{1 - {a^3}}}\\
 = \dfrac{{\left( {1 - \sqrt a } \right) + \left( {1 + \sqrt a } \right)}}{{2.\left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)}} - \dfrac{{{a^2} + 2}}{{{1^3} - {a^3}}}\\
 = \dfrac{{1 - \sqrt a  + 1 + \sqrt a }}{{2.\left( {{1^2} - {{\sqrt a }^2}} \right)}} - \dfrac{{{a^2} + 2}}{{\left( {1 - a} \right).\left( {{1^2} + 1.a + {a^2}} \right)}}\\
 = \dfrac{2}{{2\left( {1 - a} \right)}} - \dfrac{{{a^2} + 2}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
 = \dfrac{1}{{1 - a}} - \dfrac{{{a^2} + 2}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
 = \dfrac{{\left( {1 + a + {a^2}} \right) - \left( {{a^2} + 2} \right)}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
 = \dfrac{{1 + a + {a^2} - {a^2} - 2}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
 = \dfrac{{a - 1}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
 = \dfrac{{ - 1}}{{1 + a + {a^2}}}\\
a = \sqrt 2  \Rightarrow A = \dfrac{{ - 1}}{{1 + \sqrt 2  + {{\sqrt 2 }^2}}} = \dfrac{{ - 1}}{{3 + \sqrt 2 }}\\
 = \dfrac{{ - \left( {3 - \sqrt 2 } \right)}}{{\left( {3 - \sqrt 2 } \right)\left( {3 + \sqrt 2 } \right)}} = \dfrac{{ - 3 + \sqrt 2 }}{{{3^2} - {{\sqrt 2 }^2}}}\\
 = \dfrac{{\sqrt 2  - 3}}{{9 - 2}} = \dfrac{{\sqrt 2  - 3}}{7}\\
b,\\
DKXD:\,\,\,{a^2} - 1 \ge 0 \Leftrightarrow {a^2} \ge 1 \Leftrightarrow \left[ \begin{array}{l}
a \ge 1\\
a \le  - 1
\end{array} \right.\\
B = \sqrt {{a^2} + 2\sqrt {{a^2} - 1} }  - \sqrt {{a^2} - 2\sqrt {{a^2} - 1} } \\
 = \sqrt {\left( {{a^2} - 1} \right) + 2\sqrt {{a^2} - 1}  + 1}  - \sqrt {\left( {{a^2} - 1} \right) - 2\sqrt {{a^2} - 1}  + 1} \\
 = \sqrt {{{\sqrt {{a^2} - 1} }^2} + 2.\sqrt {{a^2} - 1} .1 + {1^2}}  - \sqrt {{{\sqrt {{a^2} - 1} }^2} - 2.\sqrt {{a^2} - 1} .1 + {1^2}} \\
 = \sqrt {{{\left( {\sqrt {{a^2} - 1}  + 1} \right)}^2}}  - \sqrt {{{\left( {\sqrt {{a^2} - 1}  - 1} \right)}^2}} \\
 = \left| {\sqrt {{a^2} - 1}  + 1} \right| - \left| {\sqrt {{a^2} - 1}  - 1} \right|\\
a = \sqrt 5  \Rightarrow B = \left| {\sqrt {{{\sqrt 5 }^2} - 1}  + 1} \right| - \left| {\sqrt {{{\sqrt 5 }^2} - 1}  - 1} \right|\\
 = \left| {\sqrt {5 - 1}  + 1} \right| - \left| {\sqrt {5 - 1}  - 1} \right|\\
 = \left| {\sqrt 4  + 1} \right| - \left| {\sqrt 4  - 1} \right|\\
 = \left| {2 + 1} \right| - \left| {2 - 1} \right|\\
 = \left| 3 \right| - \left| 1 \right|\\
 = 3 - 1\\
 = 2\\
2,\\
a,\\
M = \left( {\dfrac{1}{{a - \sqrt a }} + \dfrac{1}{{\sqrt a  - 1}}} \right):\dfrac{{\sqrt a  + 1}}{{a - 2\sqrt a  + 1}}\\
 = \left( {\dfrac{1}{{\sqrt a \left( {\sqrt a  - 1} \right)}} + \dfrac{1}{{\sqrt a  - 1}}} \right):\dfrac{{\sqrt a  + 1}}{{{{\sqrt a }^2} - 2.\sqrt a .1 + {1^2}}}\\
 = \dfrac{{1 + \sqrt a }}{{\sqrt a \left( {\sqrt a  - 1} \right)}}:\dfrac{{\sqrt a  + 1}}{{{{\left( {\sqrt a  - 1} \right)}^2}}}\\
 = \dfrac{{\sqrt a  + 1}}{{\sqrt a \left( {\sqrt a  - 1} \right)}}.\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{\sqrt a  + 1}}\\
 = \dfrac{{\sqrt a  - 1}}{{\sqrt a }}\\
b,\\
M = \dfrac{{\sqrt a  - 1}}{{\sqrt a }} = 1 - \dfrac{1}{{\sqrt a }}\\
\dfrac{1}{{\sqrt a }} > 0,\,\,\,\forall a > 0,a \ne 1\\
 \Rightarrow 1 - \dfrac{1}{{\sqrt a }} < 1,\,\,\,\forall a > 0,a \ne 1\\
 \Rightarrow M < 1,\,\,\,\forall a > 0,a \ne 1
\end{array}\)