\(\Delta'=\left(m-1\right)^2-m+3=m^2-3m+4=\left(m-\frac{3}{2}\right)^2+\frac{7}{4}>0\)
Phương trình luôn có 2 nghiệm pb
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m-3\end{matrix}\right.\)
Để biểu thức đề bài xác định \(\Leftrightarrow\left\{{}\begin{matrix}x_1\ge0\\x_2\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2>0\\x_1x_2\ge0\end{matrix}\right.\) \(\Rightarrow m\ge3\)
\(\sqrt{x_1}+\sqrt{x_2}=5\)
\(\Leftrightarrow x_1+x_2+2\sqrt{x_1x_2}=25\)
\(\Leftrightarrow2\left(m-1\right)+2\sqrt{m-3}=25\)
Đặt \(\sqrt{m-3}=a\ge0\Rightarrow m=a^2+3\) ta được:
\(2\left(a^2+3-1\right)+2a=25\)
\(\Leftrightarrow2a^2+2a-21=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{-1-\sqrt{43}}{2}< 0\left(l\right)\\a=\frac{-1+\sqrt{43}}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{m-3}=\frac{-1+\sqrt{43}}{2}\Rightarrow m-3=\frac{22-\sqrt{43}}{2}\Rightarrow m=\frac{28-\sqrt{43}}{2}\)