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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, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)
\(y=2-\left(-cosx\right).\left(-sinx\right)\)
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 = \(\sqrt{5}\)
\(y=2+2cos\left(x-\frac{\pi}{6}\right)-7=2cos\left(x-\frac{\pi}{6}\right)-5\)
\(0\le x\le\pi\Rightarrow-\frac{\pi}{6}\le x-\frac{\pi}{6}\le\frac{5\pi}{6}\)
\(\Rightarrow-\frac{\sqrt{3}}{2}\le cos\left(x-\frac{\pi}{6}\right)\le1\)
\(\Rightarrow-\sqrt{3}-5\le y\le-3\)
\(y_{min}=-\sqrt{3}-5\) khi \(x=\pi\)
\(y_{max}=-3\) khi \(x=\frac{\pi}{6}\)
Nguyễn Lê Phước ThịnhPhạm Vũ Trí DũngMiyuki Misaki
giúp e vs ạ
a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)
Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=3-t^2\)
\(\Rightarrow y_{min}=minf\left(t\right)=2\)
\(y_{max}=maxf\left(t\right)=3\)
\(y=4cos^2\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)-7=2\left[cos\left(x-\dfrac{\pi}{6}\right)+1\right]-7=2cos\left(x-\dfrac{\pi}{6}\right)-5\)
Đặt \(x-\dfrac{\pi}{6}=t\Rightarrow t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\)
\(\Rightarrow y=2cost-5\)
Do \(t\in\left[-\dfrac{\pi}{6};\dfrac{5\pi}{6}\right]\Rightarrow cost\in\left[-\dfrac{\sqrt{3}}{2};1\right]\)
\(\Rightarrow y\in\left[-5-\sqrt{3};-3\right]\)
\(y_{max}=-3\) khi \(t=0\) hay \(x=\dfrac{\pi}{6}\)
\(y_{min}=-5-\sqrt{3}\) khi \(y=\dfrac{5\pi}{6}\) hay \(x=\pi\)