Đáp án:
$\begin{array}{l}
1)P = \dfrac{{3\sqrt x }}{{\sqrt x + 2}} + \dfrac{5}{{\sqrt x - 2}} - \dfrac{{3x + 8}}{{x - 4}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right) + 5\left( {\sqrt x + 2} \right) - 3x - 8}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x + 5\sqrt x + 10 - 3x - 8}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - \sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 1}}{{\sqrt x + 2}}\\
2)\Delta > 0\\
\Leftrightarrow {m^2} - 4\left( {m - 1} \right) > 0\\
\Leftrightarrow {m^2} - 4m + 4 > 0\\
\Leftrightarrow {\left( {m - 2} \right)^2} > 0\\
\Leftrightarrow m\# 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
3{x_1} + 2{x_2} = 7\\
\Leftrightarrow {x_2} = \dfrac{{7 - 3{x_1}}}{2}\\
\Leftrightarrow {x_1} + \dfrac{{7 - 3{x_1}}}{2} = m\\
\Leftrightarrow - \dfrac{1}{2}{x_1} = m - \dfrac{7}{2}\\
\Leftrightarrow {x_1} = 7 - 2m\\
\Leftrightarrow {x_2} = \dfrac{{7 - 3\left( {7 - 2m} \right)}}{2} = 3m - 7\\
\Leftrightarrow \left( {7 - 2m} \right)\left( {3m - 7} \right) = m - 1\\
\Leftrightarrow 21m - 49 - 6{m^2} + 14m = m - 1\\
\Leftrightarrow 6{m^2} - 34m + 48 = 0\\
\Leftrightarrow \left( {3m - 8} \right)\left( {2m - 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
m = 3\left( {tm} \right)\\
m = \dfrac{8}{3}\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = 3;m = \dfrac{8}{3}
\end{array}$
