Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 4\\
P = \left( {\dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}} - \dfrac{{4x}}{{4 - x}}} \right)\\
:\dfrac{{x - 3\sqrt x }}{{10\sqrt x - 5x}}\\
= \left[ {\dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}} + \dfrac{4}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right]\\
:\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{5\sqrt x \left( {2 - \sqrt x } \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 2} \right)}^2} - {{\left( {\sqrt x - 2} \right)}^2} + 4x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{5\left( {2 - \sqrt x } \right)}}{{\sqrt x - 3}}\left( {x \ne 9} \right)\\
= \dfrac{{x + 4\sqrt x + 4 - x + 4\sqrt x - 4 + 4x}}{{\sqrt x + 2}}.\dfrac{{ - 5}}{{\sqrt x - 3}}\\
= \dfrac{{4x + 8\sqrt x }}{{\sqrt x + 2}}.\dfrac{{ - 5}}{{\sqrt x - 3}}\\
= \dfrac{{4\sqrt x }}{1}.\dfrac{{ - 5}}{{\sqrt x - 3}}\\
= \dfrac{{20\sqrt x }}{{3 - \sqrt x }}\\
b)Dkxd:x \ge 0;x \ne 4;x \ne 9\\
P = 0\\
\Rightarrow 20\sqrt x = 0\\
\Rightarrow x = 0\left( {tmdk} \right)\\
Vay\,x = 0\\
c)P = \dfrac{{20\sqrt x }}{{3 - \sqrt x }} = 20.\dfrac{{\sqrt x }}{{3 - \sqrt x }}\\
Do:20.\dfrac{{\sqrt x }}{{3 - \sqrt x }} \vdots 20\forall xTMDK
\end{array}$
Vậy P chia hết cho 20 với mọi số nguyên x sao cho $x \ge 0;x \ne 4;x \ne 9$
