\(\Delta=\left[-\left(m+3\right)\right]^2-8m\)
\(=m^2+6m+9-8m\)
\(=m^2-2m+9\)
\(=\left(m-1\right)^2+8\ge8>0\)
`->` pt luôn có 2 nghiệm phân biệt với mọi `m`
Theo hệ thức Vi-ét,ta có: \(\left\{{}\begin{matrix}x_1+x_2=m+3\\x_1x_2=2m\end{matrix}\right.\)
\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\) ; \(x_1;x_2\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{x_2}+\sqrt{x_1}}{\sqrt{x_1x_2}}=\dfrac{3}{2}\)
\(\Leftrightarrow2\left(\sqrt{x_1}+\sqrt{x_2}\right)=3\sqrt{x_1x_2}\)
\(\Leftrightarrow4\left(x_1+x_2+2\sqrt{x_1x_2}\right)=9x_1x_2\)
\(\Leftrightarrow4\left(m+3+2\sqrt{2m}\right)=18m\)
\(\Leftrightarrow4m+12+8\sqrt{2m}=18m\)
\(\Leftrightarrow14m-8\sqrt{2m}-12=0\)
Đặt \(\sqrt{m}=a;a\ge0\)
\(\Leftrightarrow14a^2-8\sqrt{2}a-12=0\)
\(\Delta=\left(-8\sqrt{2}\right)^2-4.14.\left(-12\right)=128+672=800>0\)
\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{8\sqrt{2}+\sqrt{800}}{28}=\sqrt{2}\left(tm\right)\\x_2=\dfrac{8\sqrt{2}-\sqrt{800}}{28}=-\dfrac{3\sqrt{2}}{7}\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{m}=\sqrt{2}\)
\(\Leftrightarrow m=2\)
Vậy \(m=2\)
