Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
1 + x \ge 0\\
1 - x \ge 0\\
1 - {x^2} \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x \le 1\\
x \ne 1;x \ne - 1
\end{array} \right.\\
\Rightarrow - 1 < x < 1\\
Vậy\, - 1 < x < 1\\
b)A = \left( {\dfrac{1}{{\sqrt {1 + x} }} + \sqrt {1 - x} } \right):\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }} + 1} \right)\\
= \dfrac{{1 + \sqrt {1 - x} .\sqrt {1 + x} }}{{\sqrt {1 + x} }}:\dfrac{{1 + \sqrt {1 - {x^2}} }}{{\sqrt {1 - {x^2}} }}\\
= \dfrac{{1 + \sqrt {1 - {x^2}} }}{{\sqrt {1 + x} }}.\dfrac{{\sqrt {1 - x} .\sqrt {1 + x} }}{{1 + \sqrt {1 - {x^2}} }}\\
= \sqrt {1 - x} \\
c)x = \dfrac{{\sqrt 3 }}{{2 + \sqrt 3 }}\\
\Rightarrow 1 - x = 1 - \dfrac{{\sqrt 3 }}{{2 + \sqrt 3 }} = \dfrac{{2 + \sqrt 3 - \sqrt 3 }}{{2 + \sqrt 3 }}\\
= \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{{2.2}}{{4 + 2\sqrt 3 }} = \dfrac{{{2^2}}}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}\\
\Rightarrow A = \sqrt {1 - x} = \sqrt {\dfrac{{{2^2}}}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}} = \dfrac{2}{{\sqrt 3 + 1}}\\
= \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} = \sqrt 3 - 1\\
Vậy\,A = \sqrt 3 - 1
\end{array}$
