Giải thích các bước giải:
\(\begin{array}{l}
a.A = \frac{{x + 5\sqrt x }}{{x - 25}} = \frac{{\sqrt x (\sqrt x + 5)}}{{(\sqrt x - 5)(\sqrt x + 5)}} = \frac{{\sqrt x }}{{\sqrt x - 5}} = 0\\
\leftrightarrow \sqrt x = 0\\
\leftrightarrow x = 0\\
b.B = \frac{{2\sqrt x }}{{\sqrt x - 3}} - \frac{{x + 9\sqrt x }}{{x - 9}}\\
= \frac{{2\sqrt x (\sqrt x + 3)}}{{(\sqrt x - 3)(\sqrt x + 3)}} - \frac{{x + 9\sqrt x }}{{(\sqrt x - 3)(\sqrt x + 3)}}\\
= \frac{{2x + 6\sqrt x - x - 9\sqrt x }}{{(\sqrt x - 3)(\sqrt x + 3)}}\\
= \frac{{x - 3\sqrt x }}{{(\sqrt x - 3)(\sqrt x + 3)}}\\
= \frac{{\sqrt x (\sqrt x - 3)}}{{(\sqrt x - 3)(\sqrt x + 3)}}\\
= \frac{{\sqrt x }}{{\sqrt x + 3}}\\
c.P = B:A = \frac{{\sqrt x }}{{\sqrt x + 3}}:\frac{{\sqrt x }}{{\sqrt x - 5}}\\
= \frac{{\sqrt x }}{{\sqrt x + 3}}.\frac{{\sqrt x - 5}}{{\sqrt x }} = \frac{{\sqrt x - 5}}{{\sqrt x + 3}}\\
= \frac{{\sqrt x + 3 - 8}}{{\sqrt x + 3}} = 1 - \frac{8}{{\sqrt x + 3}}\\
\sqrt x + 3 \ge 3 \leftrightarrow 0 < \frac{8}{{\sqrt x + 3}} \le \frac{8}{3}\\
\to P < 1
\end{array}\)
