Đáp án:
$\begin{array}{l}
1)a)Dkxd:x \ge 0;x \ne 1\\
P = \dfrac{{15\sqrt x - 11}}{{x + 2\sqrt x - 3}} + \dfrac{{3\sqrt x - 2}}{{1 - \sqrt x }} - \dfrac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\
= \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - \left( {5\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)P = \dfrac{1}{2}\\
\Rightarrow \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{2}\\
\Rightarrow \sqrt x + 3 = 4 - 10\sqrt x \\
\Rightarrow 11\sqrt x = 1\\
\Rightarrow \sqrt x = \dfrac{1}{{11}}\\
\Rightarrow x = \dfrac{1}{{121}}\left( {tmdk} \right)\\
3)a){x^2} - 2\left( {m + 1} \right)x + m - 4 = 0\\
\Rightarrow m - 4 < 0\\
\Rightarrow m < 4\\
b)\Delta ' = {\left( {m + 1} \right)^2} - m + 4\\
= {m^2} + 2m + 1 - m + 4\\
= {m^2} + m + \dfrac{1}{4} + \dfrac{{19}}{4}\\
= {\left( {m + \dfrac{1}{2}} \right)^2} + \dfrac{{19}}{4} \ge \dfrac{{19}}{4} > 0
\end{array}$
Vậy pt luôn có 2 nghiệm phân biệt với mọi m.
$\begin{array}{l}
3)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = m - 4
\end{array} \right.\\
M = {x_2}\left( {1 - {x_1}} \right) + {x_1}\left( {1 - {x_2}} \right)\\
= {x_2} + {x_1} - 2{x_1}{x_2}\\
= 2\left( {m + 1} \right) - 2\left( {m - 4} \right)\\
= 2m + 2 - 2m + 8\\
= 10
\end{array}$
Vậy M ko phụ thuộc vào m.
