Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 1;x\# 4\\
P = \left( {\dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - \dfrac{{\sqrt x }}{{\sqrt x + 2}} - \dfrac{{3\sqrt x + 10}}{{x - 4}}} \right):\dfrac{3}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x + 2} \right)}^2} - \sqrt x \left( {\sqrt x - 2} \right) - 3\sqrt x - 10}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 1}}{3}\\
= \dfrac{{x + 4\sqrt x + 4 - x + 2\sqrt x - 3\sqrt x - 10}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 1}}{3}\\
= \dfrac{{3\sqrt x - 6}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 1}}{3}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
b)P < \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} - \dfrac{1}{2} < 0\\
\Leftrightarrow \dfrac{{2\sqrt x - 2 - \sqrt x - 2}}{{2\left( {\sqrt x + 2} \right)}} < 0\\
\Leftrightarrow \sqrt x - 4 < 0\\
\Leftrightarrow \sqrt x < 4\\
\Leftrightarrow x < 16\\
Vậy\,0 \le x < 16;x\# 1;x\# 4\\
c)P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x + 2 - 3}}{{\sqrt x + 2}}\\
= 1 - \dfrac{3}{{\sqrt x + 2}}\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 = 3\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {ktm} \right)
\end{array}$
Vậy ko có x nguyên để P có giá trị nguyên
