\(\Leftrightarrow x^2-2\left(m+1\right)x+m+1=0\) (\(x\ne0\) \(\Rightarrow m\ne-1\))
\(\Delta'=\left(m+1\right)^2-\left(m+1\right)=m^2+m=0\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x_1=x_2=\frac{2\left(m+1\right)}{2}=1\)
Để pt có 2 nghiệm pb \(\Leftrightarrow m^2+m>0\Rightarrow\left[{}\begin{matrix}m>0\\m< -1\end{matrix}\right.\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m+1\end{matrix}\right.\)
Để \(x_1-x_2=4\) kết hợp Viet ta có hệ:
\(\left\{{}\begin{matrix}x_1-x_2=4\\x_1+x_2=2\left(m+1\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=m+3\\x_2=m-1\end{matrix}\right.\)
\(\Rightarrow\left(m+3\right)\left(m-1\right)=m+1\)
\(\Leftrightarrow x^2+m-4=0\Rightarrow m=\frac{-1\pm\sqrt{17}}{2}\)
c/ \(A=2\left(x_1^2+x_2^2\right)-x_1x_2=2\left(x_1+x_2\right)^2-5x_1x_2\)
\(A=8\left(m+1\right)^2-5\left(m+1\right)=8\left[\left(m+1\right)^2-\frac{5}{8}\left(m+1\right)+\frac{25}{256}\right]-\frac{25}{32}\)
\(A=8\left(m+1-\frac{5}{16}\right)^2-\frac{25}{32}=8\left(m+\frac{11}{16}\right)-\frac{25}{32}\ge-\frac{25}{32}\)
\(\Rightarrow A_{min}=-\frac{25}{32}\) khi \(m=-\frac{11}{16}\)