Đáp án:
$\begin{array}{l}
1)\dfrac{5}{{\sqrt 7 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 - 1}} + \dfrac{{7 + \sqrt 7 }}{{\sqrt 7 + 1}}\\
= \dfrac{{5\left( {\sqrt 7 - \sqrt 2 } \right)}}{{7 - 2}} + \dfrac{{\sqrt 2 + 1}}{{2 - 1}} + \dfrac{{\sqrt 7 \left( {\sqrt 7 + 1} \right)}}{{\sqrt 7 + 1}}\\
= \dfrac{{5\left( {\sqrt 7 - \sqrt 2 } \right)}}{5} + \sqrt 2 + 1 + \sqrt 7 + 1\\
= \sqrt 7 - \sqrt 2 + \sqrt 2 + \sqrt 7 + 2\\
= 2\sqrt 7 + 2\\
b){\left( {\sqrt 7 - \sqrt 2 } \right)^2} + \sqrt {56} \\
= 7 - 2\sqrt 7 .\sqrt 2 + 2 + \sqrt {4.14} \\
= 9 - 2\sqrt {14} + 2\sqrt {14} \\
= 9\\
c)\sqrt {9 - 4\sqrt 5 } + \sqrt {12 - 2\sqrt {35} } \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} + \sqrt {{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}} \\
= \sqrt 5 - 2 + \sqrt 7 - \sqrt 5 \\
= \sqrt 7 - 2\\
2)a)Dkxd:\left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.\\
b)M = \left( {\dfrac{{\sqrt a + 2}}{{a + 2\sqrt a + 1}} - \dfrac{{\sqrt a - 2}}{{a - 1}}} \right).\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \left[ {\dfrac{{\sqrt a + 2}}{{{{\left( {\sqrt a + 1} \right)}^2}}} - \dfrac{{\sqrt a - 2}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}} \right].\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a - 2} \right)\left( {\sqrt a + 1} \right)}}{{{{\left( {\sqrt a + 1} \right)}^2}.\left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
= \dfrac{{a + \sqrt a - 2 - \left( {a - \sqrt a - 2} \right)}}{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}.\dfrac{1}{{\sqrt a }}\\
= \dfrac{{2\sqrt a }}{{a - 1}}.\dfrac{1}{{\sqrt a }}\\
= \dfrac{2}{{a - 1}}\\
c)Dkxd:a > 0;a \ne 1\\
M = \dfrac{2}{{a - 1}} \in Z\\
\Rightarrow \left( {a - 1} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Rightarrow a \in \left\{ { - 1;0;2;3} \right\}\\
Do:a > 0;a \ne 1\\
\Rightarrow a \in \left\{ {2;3} \right\}
\end{array}$
