Đáp án:
$\begin{array}{l}
B1)\\
1)P = \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) + 2\sqrt x \left( {\sqrt x + 3} \right) - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{3}{{\sqrt x + 3}}\\
2)P = \dfrac{1}{3}\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} = \dfrac{1}{3}\\
\Leftrightarrow \sqrt x + 3 = 9\\
\Leftrightarrow \sqrt x = 6\\
\Leftrightarrow x = 36\left( {tm} \right)\\
Vậy\,x = 36\\
3)P = \dfrac{3}{{\sqrt x + 3}}\\
Do:\sqrt x + 3 \ge 3\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} \le \dfrac{1}{3}\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 3}} \le 1\\
\Leftrightarrow GTLN:P = 1\,khi:x = 0\\
B2)\\
1)A = \dfrac{{\sqrt x + 2}}{{\sqrt x - 5}}\\
x = 9\left( {tm} \right) \Leftrightarrow \sqrt x = 3\\
\Leftrightarrow A = \dfrac{{3 + 2}}{{3 - 5}} = \dfrac{{ - 5}}{2}\\
2)B = \dfrac{3}{{\sqrt x + 5}} + \dfrac{{20 - 2\sqrt x }}{{x - 25}}\\
= \dfrac{{3\left( {\sqrt x - 5} \right) + 20 - 2\sqrt x }}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{{\sqrt x + 5}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}}\\
= \dfrac{1}{{\sqrt x - 5}}\\
3)A = B.\left| {x - 4} \right|\\
\Leftrightarrow \dfrac{{\sqrt x + 2}}{{\sqrt x - 5}} = \dfrac{1}{{\sqrt x - 5}}.\left| {x - 4} \right|\\
\Leftrightarrow \sqrt x + 2 = \left| {x - 4} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 2 = x - 4\\
\sqrt x + 2 = - x + 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - \sqrt x - 6 = 0\\
x + \sqrt x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {\sqrt x - 3} \right)\left( {\sqrt x + 2} \right) = 0\\
\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 9\left( {tm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 1;x = 9
\end{array}$
