Tìm m để hàm số $y=x^{3}+(1-2m)x^{2}+(2-m)x+m+2$  có điểm cực đại và cực tiểu đồng thời điểm cực tiểu có hoành độ nhỏ hơn 1
 Ta có : y' = 3$x^{2}$ + 2.(1-2m).x + 2 - m
 Để hàm số có cực trị <=> $\Delta '$y' >0 <=> m > $\frac{5}{4}$ hoặc m < 1  (1)
 Theo Vi-ét: $\left\{ \begin{array}{l} x1+x2=\frac{-2}{3}.(1-2m)\\ x1.x2=\frac{2-m}{3} \end{array} \right.$
    Để điểm cực tiểu có hoành độ < 1
 <=> $\left\{ \begin{array}{l} (x1-1).(x2-1)>0\\ x1+x2<2 \end{array} \right.$
 <=> $\left\{ \begin{array}{l} m<\frac{7}{5}\\ m<2 \end{array} \right.$ (2)
 Kết hợp (1) và (2) => m $\in $ (-$\ \infty$ ;-1) $\cup $ ($\frac{5}{4}; \frac{7}{5}$)

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