Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 9
\end{array} \right.\\
D = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x - 9}}} \right):\left( {\dfrac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\\
= \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right):\dfrac{{\left( {2\sqrt x - 2} \right) - \left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - \left( {3x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3.\left( {\sqrt x + 1} \right)}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b,\\
D = - 1 \Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} = - 1\\
\Leftrightarrow \sqrt x + 3 = 3\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\\
c,\\
D \in Z \Rightarrow \dfrac{{ - 3}}{{\sqrt x + 1}} \in Z\\
x \in Z \Rightarrow \left( {\sqrt x + 1} \right) \in \left\{ { \pm 1; \pm 3} \right\}\\
\sqrt x + 1 \ge 1,\,\,\,\forall x \ge 0,x \ne 9\\
\Rightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;3} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Rightarrow x \in \left\{ {0;4} \right\}\\
d,\\
D = 16 \Leftrightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} = 16 \Leftrightarrow \sqrt x + 3 = \dfrac{{ - 3}}{{16}}\,\,\,\,\,\left( {vn,\,\,\,\sqrt x + 3 \ge 3,\,\,\forall x \ge 0} \right)\\
e,\\
D = \dfrac{{ - 3}}{{\sqrt x + 3}}\\
\sqrt x \ge 0 \Rightarrow \sqrt x + 3 \ge 3 \Rightarrow \dfrac{3}{{\sqrt x + 3}} \le \dfrac{3}{3} = 1\\
\Rightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} \ge - 1\\
\Rightarrow D \ge - 1,\,\,\,\forall x \ge 0,x \ne 9\\
\Rightarrow {D_{\min }} = - 1 \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0
\end{array}\)
