Toán 12 Gtnn

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nguyenbahiep1

[TEX]f(x) = \frac{x}{\sqrt{1-x}} + \frac{1-x}{\sqrt{x}} \\ txd: 0 < x <1[/TEX]

khảo sát và tìm ra max min

[TEX]f(x) = \frac{\sqrt{x^3}+ \sqrt{1-x}^3}{\sqrt{x(1-x)}} = \frac{(\sqrt{x}+\sqrt{1-x})(1-\sqrt{x(1-x)})}{\sqrt{x(1-x)}} \\ u = \sqrt{x} + \sqrt{(1-x)} \\ dk : 1 < u \leq \sqrt{2} \\ \Rightarrow u^2 -1 = 2.\sqrt{x(1-x)} \\ f(u) = \frac{-u^3+3u}{u^2-1}[/TEX]
 
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