Đáp án:
$\begin{array}{l}
a)x = 11 - 6\sqrt 2 \left( {tmdk} \right)\\
\Leftrightarrow x = 9 - 2.3.\sqrt 2 + 2\\
= {\left( {3 - \sqrt 2 } \right)^2}\\
\Leftrightarrow \sqrt x = 3 - \sqrt 2 \\
A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{3 - \sqrt 2 }}{{3 - \sqrt 2 + 3}}\\
= \dfrac{{3 - \sqrt 2 }}{{6 - \sqrt 2 }}\\
= \dfrac{{\left( {3 - \sqrt 2 } \right)\left( {6 + \sqrt 2 } \right)}}{{36 - 2}}\\
= \dfrac{{18 + 3\sqrt 2 - 6\sqrt 2 - 2}}{{34}}\\
= \dfrac{{16 - 3\sqrt 2 }}{{34}}\\
b)A = \dfrac{{\sqrt x }}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 - 3}}{{\sqrt x + 3}}\\
= 1 - \dfrac{3}{{\sqrt x + 3}}\\
A \in Z\\
\Leftrightarrow 1 - \dfrac{3}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 = 3\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tmdk} \right)\\
Vay\,x = 0\\
c)A = 1 - \dfrac{3}{{\sqrt x + 3}}\\
\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \le 1\\
\Leftrightarrow - \dfrac{3}{{\sqrt x + 3}} \ge - 1\\
\Leftrightarrow 1 - \dfrac{3}{{\sqrt x + 3}} \ge 0\\
\Leftrightarrow A \ge 0\\
\Leftrightarrow GTNN:A = 0\,khi:x = 0\\
d)B = \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - \left( {x - 6\sqrt x + 9} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - {{\left( {\sqrt x - 3} \right)}^2}}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - \left( {\sqrt x - 3} \right)}}{{\sqrt x + 3}}\\
= \dfrac{{3 - \sqrt x }}{{\sqrt x + 3}}
\end{array}$
