sin⁴x+cos⁴x=(sin²x+cos²x)²-2*sin²x*cos²x=1-... thế là đưa hết về hàm bậc 2 của sin2x oy nha

\(A=\left(\dfrac{1-cos2x}{2}\right)^2+2\left(\dfrac{1+cos2x}{2}\right)^2\)

\(=\dfrac{3}{4}cos^22x+\dfrac{1}{2}cos2x+\dfrac{3}{4}\)

\(A=\dfrac{1}{12}\left(3cos2x+1\right)^2+\dfrac{2}{3}\ge\dfrac{2}{3}\)

\(A_{min}=\dfrac{2}{3}\) khi \(cos2x=-\dfrac{1}{3}\)

\(A=\dfrac{3cos^22x+2cos2x-5}{4}+2=\dfrac{\left(3cos2x+5\right)\left(cos2x-1\right)}{4}+2\le2\)

\(A_{max}=2\) khi \(cos2x=1\)

1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)

 \(y=2-\left(-cosx\right).\left(-sinx\right)\)

y = 2 - sinx.cosx

y = \(2-\dfrac{1}{2}sin2x\)

Max = 2 + \(\dfrac{1}{2}\) = 2,5

Min = \(2-\dfrac{1}{2}\) = 1,5

2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)

Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)

Max = \(\sqrt{5}\)

2.

$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$

$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$

Vì: $0\leq \sin ^22x\leq 1$

$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$

Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$

3.

$0\leq |\sin x|\leq 1$

$\Rightarrow 3\geq 3-2|\sin x|\geq 1$

Vậy $y_{\min}=1; y_{\max}=3$

1.

\(y=\cos x+\cos (x-\frac{\pi}{3})=\cos x+\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(=\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\)

\(y^2=(\frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)^2\leq (\cos ^2x+\sin ^2x)(\frac{9}{4}+\frac{3}{4})\)

\(\Leftrightarrow y^2\leq 3\Rightarrow -\sqrt{3}\leq y\leq \sqrt{3}\)

Vậy $y_{\min}=-\sqrt{3}; y_{max}=\sqrt{3}$

a)\(-1\le sinx\le1\)

\(\Leftrightarrow1\ge-sinx\ge-1\)

\(\Leftrightarrow4\ge3-sinx\ge2\) \(\Leftrightarrow16\ge\left(3-sinx\right)^2\ge4\)\(\Leftrightarrow17\ge\left(3-sinx\right)^2+1\ge5\)

\(\Leftrightarrow17\ge y\ge5\)

\(y_{min}=5\Leftrightarrow sinx=1\)\(\Leftrightarrow\)\(x=\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)

\(y_{max}=17\Leftrightarrow\)\(sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)\(\left(k\in Z\right)\)

b)\(y=\left(sin^2x+cos^2x\right)^2-2.sinx^2cos^2x\)\(=1-\dfrac{1}{2}.sin^22x\)

Có \(0\le sin^22x\le1\)\(\Leftrightarrow0\ge-\dfrac{1}{2}.sin^22x\ge-\dfrac{1}{2}\)

\(\Leftrightarrow1\ge1-\dfrac{1}{2}.sin^22x\ge\dfrac{1}{2}\)\(\Leftrightarrow1\ge y\ge\dfrac{1}{2}\)

\(y_{min}=\dfrac{1}{2}\Leftrightarrow sin^22x=1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}sin2x=-1\\sin2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

\(y_{max}=1\Leftrightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\)\(\left(k\in Z\right)\)

c)\(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=1-3sin^2x.cos^2x=1-\dfrac{3}{4}.sin^22x\)

Có \(0\le sin^22x\le1\)\(\Leftrightarrow0\ge-\dfrac{3}{4}.sin^22x\ge-\dfrac{3}{4}\)

\(\Leftrightarrow1\ge1-\dfrac{3}{4}.sin^22x\ge\dfrac{1}{4}\)\(\Leftrightarrow1\ge y\ge\dfrac{1}{4}\)

​\(y_{min}=\dfrac{1}{4}\Leftrightarrow sin^22x=1\)​\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)​​

\(y_{max}=1\Leftrightarrow sin2x=0\Leftrightarrow x=\dfrac{k\pi}{2}\)\(\left(k\in Z\right)\)

Vậy...

a, Đặt \(t=sinx\left(t\in\left[-1;1\right]\right)\)

\(y=f\left(t\right)=\left(3-t\right)^2+1=t^2-6t+10\)

\(\Rightarrow min=min\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(1\right)=5\)

\(\Rightarrow max=max\left\{f\left(-1\right);f\left(1\right)\right\}=f\left(-1\right)=17\)

b, \(y=sin^4x+cos^4x=1-2sin^2x.cos^2x=1-\dfrac{1}{2}sin^22x\)
Đặt \(t=sin2x\left(t\in\left[-1;1\right]\right)\)

\(y=f\left(t\right)=1-\dfrac{1}{2}t^2\)

\(\Rightarrow min=min\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=\dfrac{1}{2}\)

\(\Rightarrow max=max\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=1\)

c, \(y=sin^6x+cos^6x\)

\(=sin^4x+cos^4x-sin^2x.cos^2x\)

\(=1-3sin^2x.cos^2x\)

\(=1-\dfrac{3}{4}sin^22x\)

Đặt \(t=sin2x\left(t\in\left[-1;1\right]\right)\)

\(y=f\left(t\right)=1-\dfrac{3}{4}t^2\)

\(\Rightarrow min=min\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=\dfrac{1}{4}\)

\(\Rightarrow max=max\left\{f\left(-1\right);f\left(0\right);f\left(1\right)\right\}=1\)

Đặt \(\sqrt{x^2+4x+5}=t\Rightarrow t\in\left[\sqrt{5};\sqrt{17}\right]\)

\(\Rightarrow y=f\left(t\right)=t^2-2t+7\)

\(-\dfrac{b}{2a}=1\notin\left[\sqrt{5};\sqrt{17}\right]\)

\(f\left(\sqrt{5}\right)=10+4\sqrt{5}\) ; \(f\left(\sqrt{17}\right)=22+4\sqrt{17}\)

\(\Rightarrow y_{min}=10+4\sqrt{5}\) ; \(y_{max}=22+4\sqrt{17}\)

f(

Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=t\Rightarrow t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

\(\Rightarrow y=t+\dfrac{t^2-1}{2}=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\)

Xét hàm \(y=f\left(t\right)=\dfrac{1}{2}t^2+t-\dfrac{1}{2}\) trên \(\left[-\sqrt{2};\sqrt{2}\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(f\left(-\sqrt{2}\right)=\dfrac{1-2\sqrt{2}}{2}\) ; \(f\left(-1\right)=-1\) ; \(f\left(\sqrt{2}\right)=\dfrac{1+2\sqrt{2}}{2}\)

\(\Rightarrow y_{min}=-1\) ; \(y_{max}=\dfrac{1+2\sqrt{2}}{2}\)

Chọn C