Đáp án:
$\begin{array}{l}
Dkxd:0 < a < 1\\
1)Q = \left( {\dfrac{{\sqrt {a + 1} }}{{\sqrt {a + 1} - \sqrt {1 - a} }} + \dfrac{{1 - a}}{{\sqrt {1 - {a^2}} - 1 + a}}} \right).\\
\left( {\sqrt {\dfrac{1}{{{a^2}}} - 1} - \dfrac{1}{a}} \right)\sqrt {{a^2} - 2a + 1} \\
= \left( {\dfrac{{\sqrt {a + 1} }}{{\sqrt {a + 1} - \sqrt {1 - a} }} + \dfrac{{1 - a}}{{\sqrt {1 - a} \left( {\sqrt {1 + a} - \sqrt {1 - a} } \right)}}} \right)\\
.\left( {\sqrt {\dfrac{{1 - {a^2}}}{{{a^2}}}} - \dfrac{1}{a}} \right).\sqrt {{{\left( {a - 1} \right)}^2}} \\
= \dfrac{{\sqrt {a + 1} + \sqrt {1 - a} }}{{\sqrt {a + 1} - \sqrt {1 - a} }}.\dfrac{{\sqrt {1 - {a^2}} - 1}}{a}.\left( {1 - a} \right)\\
= \dfrac{{{{\left( {\sqrt {a + 1} + \sqrt {1 - a} } \right)}^2}}}{{a + 1 - \left( {1 - a} \right)}}.\dfrac{{1 - {a^2} - 1}}{{a.\left( {\sqrt {1 - {a^2}} + 1} \right)}}.\left( {1 - a} \right)\\
= \dfrac{{a + 1 + 2\sqrt {a + 1} .\sqrt {1 - a} + 1 - a}}{{2a}}.\dfrac{{ - {a^2}}}{{a\left( {\sqrt {1 - {a^2}} + 1} \right)}}.\left( {1 - a} \right)\\
= \dfrac{{2 + 2\sqrt {1 - {a^2}} }}{2}.\dfrac{{ - 1}}{{\sqrt {1 - {a^2}} + 1}}.\left( {1 - a} \right)\\
= a - 1\\
2)Do:0 < a < 1\\
\Rightarrow - 1 < a - 1 < 0\\
\Rightarrow - 1 < Q < 0\\
\Rightarrow Q < {Q^2}
\end{array}$
