sến zậy

có bn là đc rồi 

\(S=\frac{-tan\left(\frac{\pi}{2}-x\right).\left[-\cos\left(x-\frac{\pi}{2}\right)\right]+\left[-\sin^3\left(\frac{\pi}{2}-x\right)\right]}{-\sin x.\cot x}\)

\(=\frac{\cos^3x-\cot x.\sin x}{\cot x.\sin x}=\frac{\cos^3x-\cos x}{\cos x}=\cos^2x-1=-\sin^2x\)

phương trình gì dợ

ta có \(B=\sqrt[5]{x^{15}.\left(2-x^5\right)^5}=\sqrt[5]{x^5.x^5.x^5\left(2-x^5\right).\left(2-x^5\right).\left(2-x^5\right).\left(2-x^5\right).\left(2-x^5\right)}\)

\(\Leftrightarrow B=\sqrt[5]{\left(\frac{3}{5}\right)^3.\frac{5}{3}x^5.\frac{5}{3}x^5.\frac{5}{3}x^5\left(2-x^5\right).\left(2-x^5\right).\left(2-x^5\right).\left(2-x^5\right).\left(2-x^5\right)}\)

\(\le\sqrt[5]{\left(\frac{3}{5}\right)^3.\left(\frac{5x^5+5\left(2-x^5\right)}{8}\right)^8}=\sqrt[5]{\left(\frac{3}{5}\right)^3.\left(\frac{5}{4}\right)^8}\)

Dâu bằng xảy ra khi \(\frac{5}{3}x^5=2-x^5\Leftrightarrow x=\sqrt[5]{\frac{3}{4}}\)

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