Đáp án:
$\begin{array}{l}
a)G = \dfrac{{ - 22 + 5\sqrt x - x}}{{x + 2\sqrt x - 15}} + \dfrac{{3\sqrt x - 1}}{{\sqrt x + 5}} - \dfrac{{\sqrt x - 5}}{{\sqrt x - 3}}\\
= \dfrac{{ - 22 + 5\sqrt x - x + \left( {3\sqrt x - 1} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - 22 + 5\sqrt x - x + 3x - 10\sqrt x + 3 - x + 25}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 5\sqrt x + 6}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 5}}\\
b)x = 5 + 2\sqrt 6 \left( {tmdk} \right)\\
= {\left( {\sqrt 2 + \sqrt 3 } \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 + \sqrt 2 \\
\Leftrightarrow G = \dfrac{{\sqrt x - 2}}{{\sqrt x + 5}}\\
= \dfrac{{\sqrt 3 + \sqrt 2 - 2}}{{\sqrt 3 + \sqrt 2 + 5}}\\
c)G = \dfrac{{\sqrt x - 2}}{{\sqrt x + 5}} = \dfrac{{\sqrt x + 5 - 7}}{{\sqrt x + 5}}\\
= 1 - \dfrac{7}{{\sqrt x + 5}}\\
G \in Z\\
\Leftrightarrow \dfrac{7}{{\sqrt x + 5}} \in Z\\
\Leftrightarrow \sqrt x + 5 = 7\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {tm} \right)\\
Vậy\,x = 4
\end{array}$
