Đáp án:
$\begin{array}{l}
I)\\
a)Dkxd:x \ge 0;x \ne 9\\
P = \left( {\frac{{2\sqrt x }}{{\sqrt x + 3}} + \frac{{\sqrt x }}{{\sqrt x - 3}} - \frac{{3x + 3}}{{x - 9}}} \right)\\
:\left( {\frac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\\
= \frac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
:\frac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\
= \frac{{2x - 6\sqrt x + x + 3\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\frac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \frac{{3x - 3\sqrt x }}{{\sqrt x + 3}}.\frac{1}{{\sqrt x + 1}}\\
= \frac{{3\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 3}}.\frac{1}{{\sqrt x + 1}}\\
= \frac{{3\sqrt x }}{{\sqrt x + 3}}\\
b)x = 4 - 2\sqrt 3 \left( {tmdk} \right)\\
\Rightarrow x = {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 3 - 1\\
P = \frac{{3\sqrt x }}{{\sqrt x + 3}} = \frac{{3\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1 + 3}} = \frac{{3\sqrt 3 - 3}}{{2 + \sqrt 3 }}\\
= 9\sqrt 3 - 15\\
II)\\
a)m = 1\\
\Rightarrow y = - x - 1\\
+ Cho:x = 0 \Rightarrow y = - 1\\
+ Cho:x = - 1 \Rightarrow y = 0\\
\Rightarrow \left( {0; - 1} \right);\left( { - 1;0} \right) \in y = - x - 1\,khi:m = 1\\
b)A\left( {2;3} \right) \in y = \left( {2m - 3} \right).x - 1\\
\Rightarrow 3 = \left( {2m - 3} \right).2 - 1\\
\Rightarrow 3 = 4m - 6 - 1\\
\Rightarrow 4m = 10\\
\Rightarrow m = \frac{5}{2}\\
c)\left\{ \begin{array}{l}
2m - 3 = 3\\
- 1 \ne 1
\end{array} \right.\\
\Rightarrow 2m = 6\\
\Rightarrow m = 3
\end{array}$
