a/ Bạn tự giải
b/ \(\Delta=\left(m-1\right)^2+24>0\Rightarrow\) pt luôn có 2 nghiệm pb
Theo Viet ta có \(\left\{{}\begin{matrix}x_1+x_2=1-m\\x_1x_2=-6\end{matrix}\right.\)
\(B=\left(x_1-3\right)\left(x_2-2\right)\left(x_1+3\right)\left(x_2+2\right)\)
\(=\left(x_1x_2-2x_1-3x_2+6\right)\left(x_1x_2+2x_1+3x_2+6\right)\)
\(=\left(-6-\left(2x_1+3x_2\right)+6\right)\left(-6+2x_1+3x_2+6\right)\)
\(=-\left(2x_1+3x_2\right)^2\le0\)
\(\Rightarrow B_{max}=0\) khi \(2x_1+3x_2=0\)
Kết hợp Viet ta có hệ: \(\left\{{}\begin{matrix}x_1+x_2=1-m\\2x_1+3x_2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=3-3m\\x_2=2m-2\end{matrix}\right.\)
Mà \(x_1x_2=-6\Rightarrow\left(3-3m\right)\left(2m-2\right)=-6\)
\(\Leftrightarrow-6\left(m-1\right)^2=-6\)
\(\Leftrightarrow\left(m-1\right)^2=1\Rightarrow\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\)
