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Vì $x\in(-1,1)$ nên $\exists t\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$ sao cho: $x=\sin t$.
Ta có: $y=\sin^6t+4(1-\sin^2t)^3=\sin^6t+4\cos^6t$.
*) Tìm Max y:
Ta có: $y=\sin^6t+4\cos^6t\le\sin^2t+4\cos^2t\le 4(\sin^2t+\cos^2t)=4$.
Dấu bằng xảy ra khi: $x=\sin t=0$.
*) Tìm Min y:
Áp dụng Bất dẳng thức Cô-si ta có:
$\sin^6t+\frac{8}{27}+\frac{8}{27}\ge\frac{4}{3}\sin^2t\Rightarrow\sin^6t+\frac{16}{27}\ge\frac{4}{3}\sin^2t$
$\cos^6t+\frac{1}{27}+\frac{1}{27}\ge\frac{1}{3}\cos^2t \Rightarrow4\cos^6t+\frac{8}{27}\ge\frac{4}{3}\cos^2t $
Từ đó suy ra: $y+\frac{24}{27}\ge\frac{4}{3}(\sin^2t+\cos^2t)=\frac{4}{3}\Rightarrow y\ge\frac{4}{9}$.
Dấu bằng xảy ra khi: $\sin^2t=\frac{2}{3}$ hay $x=\pm \sqrt{\frac{2}{3}}$
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