- Với \(m=0\) ko thỏa mãn
- Với \(m\ne0\)
\(\Rightarrow m\left(x^3-8\right)-x^2+2x=0\)
\(\Leftrightarrow m\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(mx^2+\left(2m-1\right)x+4m\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\mx^2+\left(2m-1\right)x+4m=0\left(1\right)\end{matrix}\right.\)
Pt có 3 nghiệm pb lớn hơn 1 khi (1) có 2 nghiệm pb khác 2 và lớn hơn 1
\(\Rightarrow\left\{{}\begin{matrix}\Delta=\left(2m-1\right)^2-16m^2>0\\f\left(2\right)=4m+2\left(2m-1\right)+4m\ne0\\\left(x_1-1\right)\left(x_2-1\right)>0\\\dfrac{x_1+x_2}{2}>1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-12m^2-4m+1>0\\m\ne\dfrac{1}{6}\\x_1x_2-\left(x_1+x_2\right)+1>0\\x_1+x_2>2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-12m^2-4m+1>0\\m\ne\dfrac{1}{6}\\4+\dfrac{2m-1}{m}+1>0\\-\dfrac{2m-1}{m}>2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}< m< \dfrac{1}{6}\\m\ne\dfrac{1}{6}\\\left[{}\begin{matrix}m< 0\\m>\dfrac{1}{7}\end{matrix}\right.\\0< m< \dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{7}< m< \dfrac{1}{6}\)
