Đáp án:
m=4
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt dương
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
\Delta > 0\\
5 > 0\left( {ld} \right)\\
m > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
25 - 4m > 0\\
m > 0
\end{array} \right.\\
\to \dfrac{{25}}{4} > m > 0\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 5\\
{x_1}{x_2} = m
\end{array} \right.\\
Do:2\left( {\dfrac{1}{{\sqrt {{x_1}} }} + \dfrac{1}{{\sqrt {{x_2}} }}} \right) = 3\\
\to \dfrac{1}{{\sqrt {{x_1}} }} + \dfrac{1}{{\sqrt {{x_2}} }} = \dfrac{3}{2}\\
\to \dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} + \dfrac{2}{{\sqrt {{x_1}{x_2}} }} = \dfrac{9}{4}\\
\to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} + \dfrac{2}{{\sqrt {{x_1}{x_2}} }} = \dfrac{9}{4}\\
\to \dfrac{5}{m} + \dfrac{2}{{\sqrt m }} = \dfrac{9}{4}\\
Đặt:\dfrac{1}{{\sqrt m }} = t\left( {t > 0} \right)\\
Pt \to 5{t^2} + 2t - \dfrac{9}{4} = 0\\
\to \left[ \begin{array}{l}
t = \dfrac{1}{2}\\
t = - \dfrac{9}{{10}}\left( l \right)
\end{array} \right.\\
\to \dfrac{1}{{\sqrt m }} = \dfrac{1}{2}\\
\to \dfrac{1}{m} = \dfrac{1}{4}\\
\to m = 4
\end{array}\)
