Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 1;x\# 9\\
x = 16\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 4\\
A = \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} = \dfrac{{4 + 2}}{{4 - 3}} = 6\\
b)B = \dfrac{{\sqrt x + 5}}{{\sqrt x + 1}} + \dfrac{{7 - \sqrt x }}{{x - 1}}\\
= \dfrac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 1} \right) + 7 - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 4\sqrt x - 5 + 7 - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}\\
c)P = A:B\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}:\dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x - 3}}\\
Khi:\sqrt {{P^2} - 1} = \sqrt {2P - 1} \\
Dkxd:P \ge 1\\
Khi:\sqrt {{P^2} - 1} = \sqrt {2P - 1} \\
\Leftrightarrow {P^2} - 1 = 2P - 1\\
\Leftrightarrow {P^2} - 2P = 0\\
\Leftrightarrow P\left( {P - 2} \right) = 0\\
\Leftrightarrow P = 2\left( {do:P \ge 1} \right)\\
\Leftrightarrow \dfrac{{\sqrt x - 1}}{{\sqrt x - 3}} = 2\\
\Leftrightarrow \sqrt x - 1 = 2\sqrt x - 6\\
\Leftrightarrow \sqrt x = 5\\
\Leftrightarrow x = 25\left( {tmdk} \right)\\
Vậy\,x = 25
\end{array}$
