a.
\(\left\{{}\begin{matrix}sin^4x\le sin^2x\\cos^3x\le cos^2x\end{matrix}\right.\) \(\Rightarrow y\le sin^2x+cos^2x=1\)
\(y_{max}=1\) khi \(\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(y=\left(1-cos^2x\right)^2+cos^3x=cos^4x+cos^3x-2cos^2x+1\)
\(y=\left(cosx+1\right)\left(cos^3x-2cosx+2\right)-1\ge-1\)
\(y_{min}=-1\) khi \(cosx=-1\)
b.
\(y=sin^4x.cos^2x\ge0\)
\(y_{min}=0\) khi \(sin2x=0\)
\(y=sin^4x\left(1-sin^2x\right)=\frac{1}{2}.sin^2x.sin^2x.\left(2-2sin^2x\right)\le\frac{1}{2}\left(\frac{sin^2x+sin^2x+2-2sin^2x}{3}\right)^3=\frac{4}{27}\)
\(y_{max}=\frac{4}{27}\) khi \(sin^2x=\frac{2}{3}\)
c.
\(y_{max}\) ko tồn tại
\(y=\frac{tanx}{2}+\frac{tanx}{2}+\frac{1}{tan^2x}\ge3\sqrt[3]{\frac{tan^2x}{4tan^2x}}=\frac{3}{\sqrt[3]{4}}\)
Dấu "=" xảy ra khi \(tanx=\sqrt[3]{2}\)
