nghịch biến nè

N

nguyenbahiep1

[TEX]y = \frac{x^2 -2.a.x+3.a^2}{2.a-x} \\ y' = \frac{-x^2+4.a.x -a^2}{(2a-x)^2} \\ TH_1 : \Delta' \leq 0 \Rightarrow 4a^2-a^2 = 3a^2 \leq 0 \Rightarrow a = 0 \\ TH_2 : \Delta' > 0 \Rightarrow \forall a \\ x_1 < x_2 < 1 \\ x_1+x_2 = 4a \\ x_1.x_2 = a^2\\ \Rightarrow x_1 -1 + x_2 -1 < 0 \Rightarrow 4a -2 < 0 \Rightarrow a < \frac{1}{2} \\ (x_1-1)(x_2-1) > 0 \Rightarrow a^2 -4a + 1 > 0 \\ \Rightarrow a < 2 -\sqrt{3}[/TEX]

chốt lại để thỏa mãn thì

[TEX] a < 2 -\sqrt{3}[/TEX]
 
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