\(\Delta'=m^2-2\left(m^2-2\right)=4-m^2\ge0\Rightarrow-2\le m\le2\)
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1.x_2=\dfrac{m^2-2}{2}\end{matrix}\right.\)
\(\Rightarrow P=\left|m^2-2-m-4\right|=\left|m^2-m-6\right|=\left|\left(m-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\right|\)
Do \(-2\le m\le2\Rightarrow0\le\left(m-\dfrac{1}{2}\right)^2\le\dfrac{25}{4}\)
\(\Rightarrow\left(m-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\le0\) \(\Rightarrow P=\dfrac{25}{4}-\left(m-\dfrac{1}{2}\right)^2\le\dfrac{25}{4}\)
\(\Rightarrow P_{max}=\dfrac{25}{4}\) ; dấu "=" xảy ra khi \(m=\dfrac{1}{2}\)
Lời giải:
Để pt có 2 nghiệm pb thì \(\Delta'=m^2-2(m^2-2)>0\Leftrightarrow 2> m> -2\)
Nếu $x_1,x_2$ là nghiệm của pt đã cho thì theo định lý Viete ta có:
\(\left\{\begin{matrix} x_1+x_2=-m\\ x_1x_2=\frac{m^2-2}{2}\end{matrix}\right.\)
Khi đó:
\(P=|2x_1x_2+x_1+x_2-4|=|2.\frac{m^2-2}{2}+(-m)-4|\)
\(=|m^2-m-6|=|(m-3)(m+2)|\)
\(=|m-3||m+2|=(3-m)(m+2)=m+6-m^2\) (do \(-2< m< 2\))
\(=\frac{25}{4}-(m-\frac{1}{2})^2\leq \frac{25}{4}\)
Vậy \(P_{\max}=\frac{25}{4}\Leftrightarrow m=\frac{1}{2}\)
