Đáp án: $x \in \left\{ {1;2;3;5;6} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
3)DKxd:x > 0;x\# 4\\
A = \dfrac{{2 + \sqrt x }}{{\sqrt x }}\\
B = \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
P = \dfrac{A}{B} = \dfrac{{2 + \sqrt x }}{{\sqrt x }}:\dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
= \dfrac{{\left( {2 + \sqrt x } \right)\left( {\sqrt x - 2} \right)}}{x} = \dfrac{{x - 4}}{x}\\
P.x \le \dfrac{3}{2}\left( {\sqrt x - 1} \right)\\
\Leftrightarrow \dfrac{{x - 4}}{x}.x \le \dfrac{3}{2}\sqrt x - \dfrac{3}{2}\\
\Leftrightarrow x - 4 \le \dfrac{3}{2}\sqrt x - \dfrac{3}{2}\\
\Leftrightarrow 2x - 3\sqrt x - 5 \le 0\\
\Leftrightarrow \left( {2\sqrt x - 5} \right)\left( {\sqrt x + 1} \right) \le 0\\
\Leftrightarrow 2\sqrt x - 5 \le 0\\
\Leftrightarrow \sqrt x \le \dfrac{5}{2}\\
\Leftrightarrow x \le \dfrac{{25}}{4}\\
\Leftrightarrow 0 < x \le \dfrac{{25}}{4};x\# 4\\
Do:x \in Z\\
\Leftrightarrow x \in \left\{ {1;2;3;5;6} \right\}\\
Vay\,x \in \left\{ {1;2;3;5;6} \right\}
\end{array}$
