Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {\dfrac{1}{{\sqrt x + 2}} + \dfrac{1}{{\sqrt x - 2}}} \right).\dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 2} \right) + \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x }}\\
= \dfrac{2}{{\sqrt x + 2}}\\
b,\\
P = \dfrac{2}{{\sqrt x + 2}}\\
\sqrt x \ge 0 \Rightarrow \sqrt x + 2 \ge 2 \Rightarrow P = \dfrac{2}{{\sqrt x + 2}} \le \dfrac{2}{2} = 1\\
\sqrt x + 2 \ge 2 > 0 \Rightarrow P = \dfrac{2}{{\sqrt x + 2}} > 0\\
\Rightarrow 0 < P \le 1\\
\Leftrightarrow 0 < \dfrac{{7P}}{3} \le \dfrac{7}{3}\\
\dfrac{{7P}}{3} \in Z \Rightarrow \dfrac{{7P}}{3} \in \left\{ {1;2} \right\}\\
TH1:\,\,\,\,\dfrac{{7P}}{3} = 1\\
\Leftrightarrow P = \dfrac{3}{7}\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 2}} = \dfrac{3}{7}\\
\Leftrightarrow 14 = 3\sqrt x + 6\\
\Leftrightarrow \sqrt x = \dfrac{8}{3}\\
\Leftrightarrow x = \dfrac{{64}}{9}\\
b,\\
\dfrac{{7P}}{3} = 2\\
\Leftrightarrow P = \dfrac{6}{7}\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 2}} = \dfrac{6}{7}\\
\Leftrightarrow 14 = 6\sqrt x + 12\\
\Leftrightarrow 6\sqrt x = 2\\
\Leftrightarrow \sqrt x = \dfrac{1}{3}\\
\Leftrightarrow x = \dfrac{1}{9}
\end{array}\)
