Đáp án:
$\begin{array}{l}
B1)\\
a)Dkxd:x > 0;x\# 9\\
P = \left( {\dfrac{{x - 6}}{{x + 3\sqrt x }} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x + 3}}} \right):\dfrac{{2\sqrt x - 6}}{{x + 1}}\\
= \left( {\dfrac{{x - 6}}{{\sqrt x \left( {\sqrt x + 3} \right)}} - \dfrac{1}{{\sqrt x }} + \dfrac{1}{{\sqrt x + 3}}} \right).\dfrac{{x + 1}}{{2\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 6 - \sqrt x - 3 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 3} \right)}}.\dfrac{{x + 1}}{{2\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 9}}{{2\sqrt x \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\left( {x + 1} \right)\\
= \dfrac{{x + 1}}{{2\sqrt x }}\\
b)P = 1\\
\Leftrightarrow \dfrac{{x + 1}}{{2\sqrt x }} = 1\\
\Leftrightarrow x + 1 = 2\sqrt x \\
\Leftrightarrow x - 2\sqrt x + 1 = 0\\
\Leftrightarrow {\left( {\sqrt x - 1} \right)^2} = 0\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1\\
B2)\\
a)Dkxd:x \ge 0;x\# 1\\
P = \left( {\dfrac{1}{{\sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{x - 1}}{{2\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1 - \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{2\sqrt x + 1}}\\
= \dfrac{{ - 2}}{{2\sqrt x + 1}}\\
b)P \le - 1\\
\Leftrightarrow \dfrac{{ - 2}}{{2\sqrt x + 1}} + 1 \le 0\\
\Leftrightarrow \dfrac{{ - 2 + 2\sqrt x + 1}}{{2\sqrt x + 1}} \le 0\\
\Leftrightarrow 2\sqrt x - 1 \le 0\\
\Leftrightarrow \sqrt x \le \dfrac{1}{2}\\
\Leftrightarrow x \le \dfrac{1}{4}\\
Vậy\,0 \le x \le \dfrac{1}{4}
\end{array}$
