Đáp án: $m = 4$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} - 5x + m = 0\\
\Delta > 0\\
\Leftrightarrow {5^2} - 4m > 0\\
\Leftrightarrow 4m < 25\\
\Leftrightarrow m < \frac{{25}}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 5\\
{x_1}{x_2} = m
\end{array} \right.\\
Khi:{x_1}\sqrt {{x_2}} + {x_2}\sqrt {{x_1}} = 6\\
\Leftrightarrow {x_1} > 0;{x_2} > 0\\
\Leftrightarrow {x_1}.{x_2} > 0\\
\Leftrightarrow m > 0\\
{x_1}\sqrt {{x_2}} + {x_2}\sqrt {{x_1}} = 6\\
\Leftrightarrow \sqrt {{x_1}{x_2}} .\left( {\sqrt {{x_1}} + \sqrt {{x_2}} } \right) = 6\\
\Leftrightarrow \sqrt m .\sqrt {{{\left( {\sqrt {{x_1}} + \sqrt {{x_2}} } \right)}^2}} = 6\\
\Leftrightarrow \sqrt m .\sqrt {{x_1} + {x_2} + 2\sqrt {{x_1}{x_2}} } = 6\\
\Leftrightarrow \sqrt m .\sqrt {5 + 2\sqrt m } = 6\\
\Leftrightarrow m\left( {5 + 2\sqrt m } \right) = 36\\
\Leftrightarrow 2m\sqrt m + 5m - 36 = 0\\
\Leftrightarrow \sqrt m = 2\\
\Leftrightarrow m = 4\left( {tmdk} \right)\\
Vậy\,m = 4
\end{array}$
