Đáp án:
$\begin{array}{l}
a)x = 25\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 5\\
A = \dfrac{{x + \sqrt x + 1}}{{\sqrt x - 4}} = \dfrac{{25 + 5 + 1}}{{5 - 4}} = 31\\
b)B = \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} + \dfrac{{5\sqrt x - 8}}{{2\sqrt x - x}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} + \dfrac{{5\sqrt x - 8}}{{\sqrt x \left( {2 - \sqrt x } \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right).\sqrt x - 5\sqrt x + 8}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 5\sqrt x + 8}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 6\sqrt x + 8}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 4} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x }}\\
c)P = A.B\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x - 4}}.\dfrac{{\sqrt x - 4}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
= \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
Theo\,Co - si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} = 2\\
\Rightarrow \sqrt x + 1 + \dfrac{1}{{\sqrt x }} \ge 2 + 1 = 3\\
\Rightarrow P \ge 3\\
\Rightarrow P > 2\\
Vậy\,P > 2
\end{array}$
