Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DK:\,\,\,\,\,x \ge 0\\
a,\\
A = \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} > 0,\,\,\,\forall x \ge 0\\
A = \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} = \dfrac{{\left( {\sqrt x + 2} \right) + 4}}{{\sqrt x + 2}} = 1 + \dfrac{4}{{\sqrt x + 2}}\\
A \in {Z^ + } \Leftrightarrow \dfrac{4}{{\sqrt x + 2}} \in Z \Leftrightarrow \sqrt x + 2 \in \left\{ { \pm 1;\,\, \pm 2;\,\, \pm 4} \right\}\\
\sqrt x + 2 \ge 2,\,\,\,\forall x \ge 0\\
\Rightarrow \sqrt x + 2 \in \left\{ {2;4} \right\}\\
\Rightarrow \sqrt x \in \left\{ {0;2} \right\}\\
\Rightarrow x \in \left\{ {0;4} \right\}\\
b,\\
A < \dfrac{{15}}{2}\\
\Leftrightarrow \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} - \dfrac{{15}}{2} < 0\\
\Leftrightarrow \dfrac{{2.\left( {\sqrt x + 6} \right) - 15.\left( {\sqrt x + 2} \right)}}{{2.\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow \dfrac{{ - 13\sqrt x - 18}}{{2.\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow - \dfrac{{13\sqrt x + 18}}{{2.\left( {\sqrt x + 2} \right)}} < 0,\,\,\,\,\,\,\,\forall x \ge 0\\
\Rightarrow \sqrt x \ge 0,\,\,A < \dfrac{{15}}{2}\\
c,\\
A = 2 \Leftrightarrow \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} = 2\\
\Leftrightarrow \sqrt x + 6 = 2.\left( {\sqrt x + 2} \right)\\
\Leftrightarrow \sqrt x + 6 = 2\sqrt x + 4\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4
\end{array}\)
