\(\Delta=25-4\left(m-2\right)=33-4m>0\Rightarrow m< \frac{33}{4}\)
Theo Viet, pt có 2 nghiệm thỏa: \(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=m-2\end{matrix}\right.\)
Để biểu thức đề bài có nghĩa \(\Rightarrow\left\{{}\begin{matrix}x_1>0\\x_2>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5>0\\m-2>0\end{matrix}\right.\) \(\Rightarrow2< m< \frac{33}{4}\)
Ta có:
\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\Leftrightarrow2\left(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{x_1x_2}}\right)=3\)
\(\Leftrightarrow2\left(\sqrt{x_1}+\sqrt{x_2}\right)=3\sqrt{x_1x_2}\)
\(\Leftrightarrow4\left(x_1+x_2+2\sqrt{x_1x_2}\right)=9x_1x_2\)
\(\Leftrightarrow4\left(5+2\sqrt{x_1x_2}\right)=9x_1x_2\)
Đặt \(\sqrt{x_1x_2}=a>0\) ta được:
\(4\left(5+2a\right)=9a^2\Rightarrow9a^2-8a-20=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{10}{9}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x_1x_2}=2\Rightarrow x_1x_2=4\Rightarrow m-2=4\Rightarrow m=6\)
