Đáp án:
$\begin{array}{l}
a)Dkxd:a \ge 0;b \ge 0;ab \ne 1\\
D = \left( {\dfrac{{\sqrt a + \sqrt b }}{{1 - \sqrt {ab} }} + \dfrac{{\sqrt a - \sqrt b }}{{1 + \sqrt {ab} }}} \right):\left( {1 + \dfrac{{a + b + 2ab}}{{1 - ab}}} \right)\\
= \dfrac{{\left( {\sqrt a + \sqrt b } \right)\left( {1 + \sqrt {ab} } \right) + \left( {\sqrt a - \sqrt b } \right)\left( {1 - \sqrt {ab} } \right)}}{{\left( {1 - \sqrt {ab} } \right)\left( {1 + \sqrt {ab} } \right)}}:\\
\dfrac{{1 - ab + a + b + 2ab}}{{1 - ab}}\\
= \dfrac{{\sqrt a + a\sqrt b + \sqrt b + b\sqrt a + \sqrt a - a\sqrt b - \sqrt b + b\sqrt a }}{{1 - ab}}.\\
\dfrac{{1 - ab}}{{a + b + ab + 1}}\\
= \dfrac{{2\sqrt a + 2b\sqrt a }}{{a + b + ab + 1}}\\
= \dfrac{{2\sqrt a \left( {b + 1} \right)}}{{\left( {b + 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{2\sqrt a }}{{a + 1}}\\
b)a = \dfrac{2}{{2 + \sqrt 3 }}\left( {tmdk} \right)\\
a = \dfrac{{2\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = 4 - 2\sqrt 3 \\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \sqrt a = \sqrt 3 - 1\\
D = \dfrac{{2\sqrt a }}{{a + 1}}\\
= \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{4 - 2\sqrt 3 + 1}}\\
= \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{5 - 2\sqrt 3 }}\\
= \dfrac{{2\left( {\sqrt 3 - 1} \right)\left( {5 + 2\sqrt 3 } \right)}}{{25 - 12}}\\
= \dfrac{{\left( {2\sqrt 3 - 2} \right)\left( {5 + 2\sqrt 3 } \right)}}{{13}}\\
= \dfrac{{10\sqrt 3 + 12 - 10 - 4\sqrt 3 }}{{13}}\\
= \dfrac{{6\sqrt 3 + 2}}{{13}}\\
c)D = \dfrac{{2\sqrt a }}{{a + 1}}\\
D.\left( {a + 1} \right) = 2\sqrt a \\
\Rightarrow D.a - 2\sqrt a + D = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow 1 - {D^2} \ge 0\\
\Rightarrow {D^2} \le 1\\
\Rightarrow 0 \le D \le 1\\
\Rightarrow GTLN:D = 1\,khi:a = 1
\end{array}$
