Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}} - \dfrac{{3\sqrt x + 6}}{{x - 4}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}} - \dfrac{{3\sqrt x + 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + {{\left( {\sqrt x - 2} \right)}^2} - \left( {3\sqrt x + 6} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {x + 3\sqrt x + 2} \right) + \left( {x - 4\sqrt x + 4} \right) - \left( {3\sqrt x + 6} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2x - 4\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x + 2}}\\
2,\\
A = \dfrac{1}{2} \Leftrightarrow \dfrac{{2\sqrt x }}{{\sqrt x + 2}} = \dfrac{1}{2}\\
\Leftrightarrow 4\sqrt x = \sqrt x + 2\\
\Leftrightarrow 3\sqrt x = 2\\
\Leftrightarrow \sqrt x = \dfrac{2}{3}\\
\Leftrightarrow x = \dfrac{4}{9}\\
3,\\
x = 16 \Rightarrow A = \dfrac{{2.\sqrt {16} }}{{\sqrt {16} + 2}} = \dfrac{{2.4}}{{4 + 2}} = \dfrac{4}{3}\\
4,\\
A = \dfrac{{2\sqrt x }}{{\sqrt x + 2}} = \dfrac{{\left( {2\sqrt x + 4} \right) - 4}}{{\sqrt x + 2}} = \dfrac{{2.\left( {\sqrt x + 2} \right) - 4}}{{\sqrt x + 2}} = 2 - \dfrac{4}{{\sqrt x + 2}}\\
A \in Z \Leftrightarrow \dfrac{4}{{\sqrt x + 2}} \in Z \Rightarrow \left( {\sqrt x + 2} \right) \in \left\{ { \pm 1; \pm 2; \pm 4} \right\}\\
\sqrt x + 2 \ge 2,\,\,\forall x \ge 0\\
\Rightarrow \left( {\sqrt x + 2} \right) \in \left\{ {2;4} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Rightarrow x \in \left\{ {0;4} \right\}\\
x \ge 0,x \ne 4 \Rightarrow x = 0
\end{array}\)
