\(\Delta'=\left[-\left(m+1\right)\right]^2+m^2=\left(m+1\right)^2+m^2>0\)
=> PT có hai nghiệm phân biệt
Theo hệ thức Vi-ét, ta có
\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m\end{matrix}\right.\)
Ta có \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(2m+2\right)^2-2m\)
\(=4m^2-8m+4-2m=4m^2-10m+4\)
Mà \(x_1^2+x_2^2=4\sqrt{x_1x_1}\)
\(\Rightarrow4m^2-10m+4=4\sqrt{m}\)
\(\Rightarrow4m^2+10m-4\sqrt{m}-4=0\)
\(\Rightarrow2\left(2m^2+5m-2\sqrt{m}-1\right)=0\)
\(\Rightarrow2m^2+5m-2\sqrt{m}-1=0\)
\(\Rightarrow2\left(m^2+2m-1\right)+\left(m-2\sqrt{m}+1\right)=0\)
\(\Rightarrow2\left(m-1\right)^2+\left(\sqrt{m}-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(m-1\right)^2=0\\\sqrt{m}-1=0\end{matrix}\right.\) => m = 1
Vậy giá trị m thỏa mãn là m= 1