\(\Delta'=\left(m-1\right)^2\ge0;\forall m\)
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=4m\end{matrix}\right.\)
a/ Để pt có nghiệm này gấp đôi nghiệm kia \(\Leftrightarrow x_1=2x_2\)
Kết hợp Viet ta được: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1=2x_2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_2=2\left(m+1\right)\\x_1=2x_2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_2=\frac{2}{3}\left(m+1\right)\\x_1=\frac{4}{3}\left(m+1\right)\end{matrix}\right.\)
Mà \(x_1x_2=4m\Leftrightarrow\frac{8}{9}\left(m+1\right)^2=4m\)
\(\Leftrightarrow2m^2+4m+2=9m\Leftrightarrow2m^2-5m+2=0\Rightarrow\left[{}\begin{matrix}m=2\\m=\frac{1}{2}\end{matrix}\right.\)
b/ Để pt có 2 nghiệm trái dấu \(\Leftrightarrow x_1x_2< 0\Leftrightarrow4m< 0\Rightarrow m< 0\)
c/ Để pt có 2 nghiệm cùng dương
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)>0\\x_1x_2=4m>0\end{matrix}\right.\) \(\Rightarrow m>0\)
d/ \(A=2\left(x_1+x_2\right)^2-5x_1x_2\)
\(A=8\left(m+1\right)^2-20m\)
\(A=8m^2-4m+8=8\left(m-\frac{1}{4}\right)^2+\frac{15}{2}\ge\frac{15}{2}\)
\(A_{min}=\frac{15}{2}\) khi \(m=\frac{1}{4}\)