Đáp án:
$\begin{array}{l}
a)Đkxđ:\left\{ \begin{array}{l}
a + 1 > 0\\
1 - a \ge 0\\
1 - {a^2} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a > - 1\\
a \le 1\\
{a^2} < 1
\end{array} \right. \Rightarrow \left\{ { - 1 < a < 1} \right.\\
b) Đkxđ: - 1 < a < 1\\
Q = \left( {\frac{1}{{\sqrt {1 + a} }} + \sqrt {1 - a} } \right):\left( {\frac{1}{{\sqrt {1 - {a^2}} }} + 1} \right)\\
= \left( {\frac{{1 + \sqrt {1 - a} .\sqrt {1 + a} }}{{\sqrt {1 + a} }}} \right):\left( {\frac{{1 + \sqrt {1 - {a^2}} }}{{\sqrt {1 - a} .\sqrt {1 + a} }}} \right)\\
= \frac{{1 + \sqrt {1 - {a^2}} }}{{\sqrt {1 + a} }}.\frac{{\sqrt {1 - a} .\sqrt {1 + a} }}{{1 + \sqrt {1 - {a^2}} }}\\
= \sqrt {1 - a}
\end{array}$
