\(f\left(mx^2-mx+2\right)=0\Rightarrow\left[{}\begin{matrix}mx^2-mx+2=0\\mx^2-mx+2=3\\mx^2-mx+2=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}mx^2-mx+2=0\\mx^2-mx-1=0\\mx^2-mx-4=0\end{matrix}\right.\)
Để pt có 6 nghiệm pb
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta_1=m^2-8m>0\\\Delta_2=m^2+4m>0\\\Delta_3=m^2+16m>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< 0\\m>8\end{matrix}\right.\\\left[{}\begin{matrix}m< -4\\m>0\end{matrix}\right.\\\left[{}\begin{matrix}m< -16\\m>0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m>8\\m< -16\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}9\le m\le2019\\-2019\le m\le-17\end{matrix}\right.\)
\(\Rightarrow\) Có \(\left(-17-\left(-2019\right)+1\right)+\left(2019-9+1\right)=4014\) giá trị
