\(\Delta'=\left(m+1\right)^2-\left(2m-3\right)=m^2+4>0;\forall m\)
Để pt có 2 nghiệm không âm: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)>0\\x_1x_2=2m-3\ge0\end{matrix}\right.\) \(\Rightarrow m\ge\frac{3}{2}\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m-3\end{matrix}\right.\)
\(\sqrt{x_1}+\sqrt{x_2}=3\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=9\)
\(\Leftrightarrow2\left(m+1\right)+2\sqrt{2m-3}=9\)
\(\Leftrightarrow2m-3+2\sqrt{2m-3}-4=0\)
Đặt \(\sqrt{2m-3}=t\ge0\)
\(\Rightarrow t^2+2t-4=0\Rightarrow\left[{}\begin{matrix}t=\sqrt{5}-1\\t=-\sqrt{5}-1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2m-3}=\sqrt{5}-1\)
\(\Leftrightarrow2m-3=6-2\sqrt{5}\Rightarrow m=\frac{9-2\sqrt{5}}{2}\)
