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P= $\frac{2cosx}{2sin^3x + 3cos^3x}$

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#1
beflower

beflower

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Cho cotx=2 Tính P= $\frac{2cosx}{2sin^3x + 3cos^3x}$



#2
bvd

bvd

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Có $\cot x = 2$ nên $\tan x = \frac{1}{2}$ và $P \ne 0$

Có $\frac{1}{P}=\tan x \sin^2x+\frac{3}{2}\cos^2x=\tan x \frac{1}{\cot^2 x+1}+\frac{3}{2}\frac{1}{\tan^2 x + 1}=\frac{13}{10}$

do đó $P=\frac{10}{13}$






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