Đáp án:
$\begin{array}{l}
x \ge 0;x \ne 1\\
a)Thay\,x = 25\,vào\,A\,ta\,đượcc:\\
A = \frac{{2\sqrt {25} - 1}}{{\sqrt {25} - 1}} = \frac{{2\sqrt {{5^2}} - 1}}{{\sqrt {{5^2}} - 1}} = \frac{{2.5 - 1}}{{5 - 1}} = \frac{{10 - 1}}{4} = \frac{9}{4}\\
b)B = \frac{{\sqrt x }}{{\sqrt x - 1}} - \frac{3}{{\sqrt x + 1}} - \frac{{6\sqrt x - 4}}{{x - 1}}\\
= \frac{{\sqrt x }}{{\sqrt x - 1}} - \frac{3}{{\sqrt x + 1}} - \frac{{6\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{\sqrt x \left( {\sqrt x + 1} \right) - 3\left( {\sqrt x - 1} \right) - 6\sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{x + \sqrt x - 3\sqrt x + 3 - 6\sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{x - 8\sqrt x + 7}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 7} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{\sqrt x - 7}}{{\sqrt x + 1}}\\
c)P = A.B = \frac{{2\sqrt x - 1}}{{\sqrt x - 1}}.\frac{{\sqrt x - 7}}{{\sqrt x + 1}} = \frac{{\left( {2\sqrt x - 1} \right).\left( {\sqrt x - 7} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \frac{{2x - 15\sqrt x + 7}}{{x - 1}}\\
P < 1\\
\Rightarrow \frac{{2x - 15\sqrt x + 7}}{{x - 1}} < 1\\
\Rightarrow \frac{{2x - 15\sqrt x + 7 - x + 1}}{{x - 1}} < 0\\
\Rightarrow \frac{{x - 15\sqrt x + 8}}{{x - 1}} < 0
\end{array}$
