Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
1.
\(y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)
\(=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x\)
Vì \(\sin 2x\in [-1;1]\Rightarrow \sin ^22x\in [0;1]\)
Do đó:\(y=1-\frac{1}{2}\sin ^22x\in [\frac{1}{2}; 1]\) hay \(y_{\min}=\frac{1}{2}; y_{\max}=1\)
2.
\(y=\frac{\sin x}{\cos x+2}\Rightarrow y^2=\frac{\sin ^2x}{(\cos x+2)^2}=\frac{1-\cos ^2x}{(\cos x+2)^2}\)
Đặt \(\cos x=t(t\in [-1;1])\) . Xét \(f(t)=\frac{1-t^2}{(t+2)^2}\)
\(f'(t)=\frac{-2(2t+1)}{(t+2)^3}=0\Leftrightarrow t=-\frac{1}{2}\)
Lập BBT ta suy ra \(f(t)_{\max}=f(\frac{-1}{2})=\frac{1}{3}\)
\(\Rightarrow y^2\leq \frac{1}{3}\Rightarrow \frac{-1}{\sqrt{3}}\leq y\leq \frac{1}{\sqrt{3}}\)
Vậy \(y_{\min}=\frac{-1}{\sqrt{3}}; y_{\max}=\frac{1}{\sqrt{3}}\)
1.
\(y=\sqrt[4]{sinx}-\sqrt{cosx}\le\sqrt[4]{sinx}\le1\)
\(y_{max}=1\) khi \(\left\{{}\begin{matrix}sinx=1\\cosx=0\end{matrix}\right.\) \(\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)
\(y=\sqrt[4]{sinx}-\sqrt{cosx}\ge-\sqrt{cosx}\ge-1\)
\(y_{min}=-1\) khi \(x=k2\pi\)
2.
\(y_{max}\) ko tồn tại
\(y=\frac{1}{cos^4x}+\frac{\sqrt{2}^2}{1-cos^4x}\ge\frac{\left(1+\sqrt{2}\right)^2}{cos^4x+1-cos^4x}=3+2\sqrt{2}\)
\(y_{min}=3+2\sqrt{2}\) khi \(cos^4x=\sqrt{2}-1\)
e/
Đề câu này chắc chắn đúng chứ bạn?
f/
\(sin^4x+cos^4x=\frac{3}{4}\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)
\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)
\(\Leftrightarrow1-2sin^22x=0\)
\(\Leftrightarrow cos4x=0\)
\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)
c/
\(y=sin\left(4x-\frac{\pi}{3}\right)+sin\left(\frac{\pi}{3}\right)+5\)
\(=sin\left(4x-\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}+5\)
Do \(-1\le sin\left(4x-\frac{\pi}{3}\right)\le1\)
\(\Rightarrow4+\frac{\sqrt{3}}{2}\le y\le6+\frac{\sqrt{3}}{2}\)
d/
\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+3sin2x+5\)
\(y=6-3sin^2x.cos^2x+3sin2x\)
\(y=-\frac{3}{4}sin^22x+3sin2x+6\)
\(y=\frac{3}{4}\left(sin2x+1\right)\left(5-sin2x\right)+\frac{9}{4}\ge\frac{9}{4}\)
\(y_{min}=\frac{9}{4}\) khi \(sin2x=-1\)
\(y=\frac{3}{4}\left(sin2x-1\right)\left(3-sin2x\right)+\frac{33}{4}\le\frac{33}{4}\)
\(y_{max}=\frac{33}{4}\) khi \(sin2x=1\)