ai giúp dùm E với

Question

$tan a=2$

Tính giá trị biểu thức:

$\frac{\sin a+6cos^{3}a+cosa}{cosa+2sin^{3}a}$

Minh................ · Answer 1 · 5/7/2016 5:16:16 PM

$\frac{\frac{sina}{cos^{3}a}+6+\frac{1}{cos^{2}a}}{\frac{1}{cos^{2}a}+2tan^{3}a}$

=$\frac{tana(tan^{2}a+1)+6+tan^{2}a+1}{tan^{2}a+1+2tan^{3}a}$

=$\frac{2(2^{2}+1)+6+2^{2}+1}{2^{2}+1+2.2^{3}}$

=1

Trả lời 07-05-16 05:16 PM

Minh................
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[_đéo_có_tên_] · Answer 2 · 5/7/2016 5:10:38 PM

$A=\frac{6cos^{3}a+(sina+ cosa)(sin^{2}a+cosa^{2})}{2sin^{3}a+cosa(sin^{2}a+cos^{2}a)}$

$A=\frac{7cos^{3}a+cos^{2}a.sina+cosa.sin^{2}a+sin^{3}a}{2sin^{3}a+sin^{2}a.cosa+cos^{3}a}$

$A=\frac{7+tana+tan^{2}a+tan^{3}a}{2tan^{3}a+tan^{2}a+1}$ ( chia hai vế cho $cos^{3}a$ )

$A=1$

Trả lời 07-05-16 05:10 PM

[_đéo_có_tên_]
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