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cho y=(mx2+6x-2)/(x+2)
tìm m để hàm số nghịch biến trên [1;+vô cực)
[TEX] y = \frac{m.x^2 + 6x -2}{(x+2)} \\ y' = \frac{m.x^2 + 4mx + 14}{(x+2)^2}\\ TH_1 : m < 0 \\ \Delta' < 0 \Rightarrow 4m^2 - 14m < 0 \Rightarrow 0 < m < \frac{7}{2} \\ \Rightarrow ( vo ly)[/TEX]
[TEX]TH_2 : m < 0 \\ \Delta' > 0 \Rightarrow m > \frac{7}{2} , m < 0 \\ x_1 < x_2 \leq 1 \\ (x_1 -1) + (x_2 -1) < 0 \\ (x_1-1)(x_2-1) \geq 0 \\ x_1+x_2 = -4 \\ x_1.x_2 = \frac{14}{m} \\ (x_1 -1) + (x_2 -1) < 0 \Rightarrow -4 - 2 < 0 \forall m \\ (x_1-1)(x_2-1) \geq 0 \Rightarrow x_1.x_2 - (x_1+x_2) +1 \geq 0 \Rightarrow \frac{14}{m} + 4 + 1 \geq 0 \Rightarrow m \leq \frac{-14}{5}[/TEX]
kết luận
[TEX]m \leq \frac{-14}{5}[/TEX]
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