Đáp án:
$\begin{array}{l}
a)x = 16\left( {tm} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow A = \frac{{4 - 5}}{4} = - \frac{1}{4}\\
b)B = \frac{{\sqrt x }}{{\sqrt x - 5}} - \frac{{3\sqrt x }}{{x - 25}}\\
= \frac{{\sqrt x \left( {\sqrt x + 5} \right) - 3\sqrt x }}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \frac{{x + 2\sqrt x }}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
P = A.B\\
= \frac{{\sqrt x - 5}}{{\sqrt x }}.\frac{{x + 2\sqrt x }}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}\\
= \frac{{\sqrt x .\left( {\sqrt x + 2} \right)}}{{\sqrt x .\left( {\sqrt x + 5} \right)}}\\
= \frac{{\sqrt x + 2}}{{\sqrt x + 5}}\\
c)P = \frac{{\sqrt x + 2}}{{\sqrt x + 5}} = 1 - \frac{3}{{\sqrt x + 5}} < 1\\
\Rightarrow P > {P^2}\\
2)a)5\sqrt {4x - 8} - 3\sqrt {9x - 18} = 4\left( {dkx \ge 2} \right)\\
\Rightarrow 5.2\sqrt {x - 2} - 3.3\sqrt {x - 2} = 4\\
\Rightarrow \sqrt {x - 2} = 4\\
\Rightarrow x - 2 = 16\\
\Rightarrow x = 18\left( {tm} \right)\\
b)\sqrt {x - 1} - \sqrt {x - 4} = 1\left( {dk:x \ge 4} \right)\\
\Rightarrow \sqrt {x - 1} = \sqrt {x - 4} + 1\\
\Rightarrow x - 1 = x - 4 + 2\sqrt {x - 4} + 1\\
\Rightarrow 2\sqrt {x - 4} = 2\\
\Rightarrow \sqrt {x - 4} = 1\\
\Rightarrow x - 4 = 1\\
\Rightarrow x = 5\left( {tm} \right)
\end{array}$
