\(\text{Δ}=\left(m+4\right)^2-8\left(m-5\right)\)
\(=m^2+8m+16-8m+40\)
\(=m^2+56>0\)
Do đó: Phương trình có hai nghiệm phân biệt
Theo đề, ta có:
\(2\left(x_1+x_2\right)^2-4x_1x_2-3\left(x_1+x_2\right)-\dfrac{1}{2}m< =0\)
\(\Leftrightarrow2\cdot\dfrac{\left(m+4\right)^2}{4}-4\cdot\dfrac{m-5}{2}-3\cdot\dfrac{m+4}{2}-\dfrac{1}{2}m< =0\)
\(\Leftrightarrow\dfrac{1}{2}m^2+4m+8-2m+10-\dfrac{3}{2}m-6-\dfrac{1}{2}m< =0\)
\(\Leftrightarrow m^2\cdot\dfrac{1}{2}+12< =0\)(vô lý)
\(\Delta=\left(m-4\right)^2-8\left(m-5\right)=m^2+56>0\)
Vậy pt có 2 nghiệm pb
Theo Vi et : \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{m+4}{2}\\x_1x_2=\dfrac{m-5}{2}\end{matrix}\right.\)
Ta có : \(2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\le3\left(x_1+x_2\right)+\dfrac{m}{2}\)
\(\Leftrightarrow2.\dfrac{\left(m+4\right)^2}{4}-4.\dfrac{m-5}{2}-3.\dfrac{m+4}{2}-\dfrac{m}{2}\le0\)
\(\Leftrightarrow\dfrac{2m^2+16m+32}{4}-\dfrac{4m-20}{2}-\dfrac{3m+12}{2}-\dfrac{m}{2}\le0\)
\(\Leftrightarrow\dfrac{2m^2+16m+32}{4}+16\le0\Leftrightarrow\dfrac{2m^2+16m+96}{4}\le0\)
\(\Rightarrow m^2+8m+48\le0\)
