Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{x + 9\sqrt x }}{{x - 9}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{x + 9\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) - \left( {x + 9\sqrt x } \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {2x + 6\sqrt x } \right) - \left( {x + 9\sqrt x } \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
B = \dfrac{{x - 5\sqrt x }}{{x - 25}} = \dfrac{{\sqrt x \left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x + 5}}\\
M = \dfrac{A}{B} = \dfrac{{\sqrt x }}{{\sqrt x + 3}}:\dfrac{{\sqrt x }}{{\sqrt x + 5}} = \dfrac{{\sqrt x + 5}}{{\sqrt x + 3}}\\
M - 1 = \dfrac{{\sqrt x + 5}}{{\sqrt x + 3}} - 1 = \dfrac{{\left( {\sqrt x + 5} \right) - \left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}} = \dfrac{2}{{\sqrt x + 3}} > 0\\
\Rightarrow M > 1\\
d,\\
B = \dfrac{{\sqrt x }}{{\sqrt x + 5}} = \dfrac{{\left( {\sqrt x + 5} \right) - 5}}{{\sqrt x + 5}} = 1 - \dfrac{5}{{\sqrt x + 5}}\\
B \in Z \Rightarrow \dfrac{5}{{\sqrt x + 5}} \in Z \Rightarrow \left( {\sqrt x + 5} \right) \in \left\{ { \pm 1; \pm 5} \right\}\\
\sqrt x \ge 0 \Rightarrow \sqrt x + 5 \ge 5 \Rightarrow \sqrt x + 5 = 5 \Rightarrow x = 0
\end{array}\)
