Tìm Max: $y=sin^4 x.cos^6 x$
Tìm Max: $y=sin^4 x.cos^6 x$
Started By zzhanamjchjzz, 24-08-2015 - 13:29
#2
Posted 24-08-2015 - 13:38
VuHongQuan
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$y=\sin^2x.\sin^2x.\cos^2x.\cos^2x.\cos^2x\\=\frac{1}{72}.(3\sin^2x)^2.(2\cos^2x)^3\le\frac{1}{72}.\frac{(6(\sin^2x+\cos^2x))^5}{5^5}=\frac{108}{3125}$
(sử dụng cô si 5 số )
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