Đáp án:
1) Phương trình có 2 nghiệm thì:
$\begin{array}{l}
\Delta ' \ge 0\\
\Rightarrow {\left( {2m - 1} \right)^2} - 4m + 3 \ge 0\\
\Rightarrow 4{m^2} - 4m + 1 - 4m + 3 \ge 0\\
\Rightarrow 4{m^2} - 8m + 4 \ge 0\\
\Rightarrow {m^2} - 2m + 1 \ge 0\\
\Rightarrow {\left( {m - 1} \right)^2} \ge 0\left( {dung\,\forall m} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {2m - 1} \right)\\
{x_1}{x_2} = 4m - 3
\end{array} \right.\\
{x_1} - {x_2} = 4\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 16\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 16\\
\Rightarrow 4{\left( {2m - 1} \right)^2} - 4.\left( {4m - 3} \right) = 16\\
\Rightarrow 4{m^2} - 4m + 1 - 4m + 3 = 4\\
\Rightarrow 4{m^2} - 8m = 0\\
\Rightarrow 4m\left( {m - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 0\\
m = 2
\end{array} \right.\\
2)Dkxd:x \ge 0\\
A = \frac{4}{{\sqrt x + 3}};B = \frac{{\sqrt x }}{{\sqrt x + 3}}\\
\Rightarrow Q = A.B\\
= \frac{{4\sqrt x }}{{{{\left( {\sqrt x + 3} \right)}^2}}} = \frac{{4\sqrt x }}{{x + 6\sqrt x + 9}}\\
\Rightarrow Q.x + 6Q.\sqrt x + 9Q = 4\sqrt x \\
\Rightarrow Q.x + 2\left( {3Q - 2} \right).\sqrt x + 9Q = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow \left( {3Q - 2} \right) - 9{Q^2} \ge 0\\
\Rightarrow \left\{ { - 12Q + 4 \ge 0} \right.\\
\Rightarrow \left\{ {Q \le \frac{1}{3}} \right.\\
\Rightarrow GTLN:Q = \frac{1}{3} \Leftrightarrow \sqrt x = 3\\
\Rightarrow x = 9\left( {tmdk} \right)
\end{array}$
