Đáp án:
1) $A=2$
2) a) $M = \dfrac{{\sqrt a - 1}}{{\sqrt a }}$ khi $DKXD:a > 0;a \ne 1$
b) $M = - \sqrt 2 $ khi $a = 3 - 2\sqrt 2 $
c)$a \in \left( {0;4} \right);a \ne 1$
Giải thích các bước giải:
$\begin{array}{l}
1)A = \left( {\sqrt {10} + \sqrt 6 } \right).\sqrt {4 - \sqrt {15} } \\
= \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt 2 .\sqrt {4 - \sqrt {15} } \\
= \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt {8 - 2\sqrt {15} } \\
= \left( {\sqrt 5 + \sqrt 3 } \right).\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \left( {\sqrt 5 + \sqrt 3 } \right).\left| {\sqrt 5 - \sqrt 3 } \right|\\
= \left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)\\
= {\left( {\sqrt 5 } \right)^2} - {\left( {\sqrt 3 } \right)^2}\\
= 2
\end{array}$
2) $DKXD:a > 0;a \ne 1$
a) Ta có:
$\begin{array}{l}
M = \left( {\dfrac{1}{{\sqrt a - 1}} + \dfrac{1}{{a - \sqrt a }}} \right):\dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \left( {\dfrac{1}{{\sqrt a - 1}} + \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right):\dfrac{{\sqrt a + 1}}{{{{\left( {\sqrt a - 1} \right)}^2}}}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a }}
\end{array}$
Vây $M = \dfrac{{\sqrt a - 1}}{{\sqrt a }}$
b) Ta có:
$\begin{array}{l}
a = 3 - 2\sqrt 2 \left( {tm:DKXD} \right)\\
\Rightarrow \sqrt a = \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = \left| {\sqrt 2 - 1} \right| = \sqrt 2
\end{array}$
Khi đó:
Với $a = 3 - 2\sqrt 2 $ thì $M = \dfrac{{\left( {\sqrt 2 - 1} \right) - 1}}{{\sqrt 2 - 1}} = \dfrac{{\sqrt 2 - 2}}{{\sqrt 2 - 1}} = - \sqrt 2 $
Vậy $M = - \sqrt 2 $ khi $a = 3 - 2\sqrt 2 $
c) Ta có:
$\begin{array}{l}
M < \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt a - 1}}{{\sqrt a }} < \dfrac{1}{2}\\
\Leftrightarrow 2\left( {\sqrt a - 1} \right) < \sqrt a \\
\Leftrightarrow \sqrt a < 2\\
\Leftrightarrow a < 4
\end{array}$
Vậy $a \in \left( {0;4} \right);a \ne 1$ thỏa mãn đề
