\(ac=-2\sqrt{2}< 0\) nên pt luôn có 2 nghiệm pb trái dấu với mọi m
Theo hệ thức Viet \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{3m}{2}\\x_1x_2=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(M=\left(x_1-x_2\right)^2+\left(\dfrac{x_2+x_1^2x_2-x_1-x_1x_2^2}{x_1x_2}\right)^2\)
\(=\left(x_1-x_2\right)^2+\left(\dfrac{x_2-x_1-x_1x_2\left(x_2-x_1\right)}{x_1x_2}\right)^2\)
\(=\left(x_1-x_2\right)^2+\left(\dfrac{\left(x_2-x_1\right)\left(1-x_1x_2\right)}{x_1x_2}\right)^2\)
\(=\left(x_1-x_2\right)^2+\dfrac{\left(x_1-x_2\right)^2\left(1-x_1x_2\right)^2}{\left(x_1x_2\right)^2}\)
\(=\left(x_1-x_2\right)^2\left(1+\left(\dfrac{1-x_1x_2}{x_1x_2}\right)^2\right)\)
\(=\left[\left(x_1+x_2\right)^2-4x_1x_2\right]\left(1+\left(\dfrac{1-x_1x_2}{x_1x_2}\right)^2\right)\)
\(=\left(\dfrac{9m^2}{4}+2\sqrt{2}\right).\left(4+2\sqrt{2}\right)\ge\left(0+2\sqrt{2}\right)\left(4+2\sqrt{2}\right)=8+8\sqrt{2}\)
Dấu "=" xảy ra khi \(m=0\)
