\(y=\dfrac{1}{3}(1-m)x^3-2(2-m)x^2+2(2-m)x+1\)
TXĐ: \(D=\mathbb R\)
\(y'=(1-m)x^2-4(2-m)x+2(2-m)\)
Để hàm số luôn nghịch biến thì \(\left\{ \begin{array}{l} {\Delta}'<0 \\ a<0 \end{array} \right .\)
\(\Rightarrow \left\{ \begin{array}{l} 4(2-m)^2-(1-m)2(2-m)<0 \\ 1-m<0 \end{array} \right .\)
\(\Rightarrow \left\{ \begin{array}{l} 2m^2-10m+12<0 \\ m>1\end{array} \right .\)
\(\Rightarrow \left\{ \begin{array}{l} 21\end{array} \right .\)
\(\Rightarrow 2
