e/ Tử số đến đâu và mẫu số đến đâu bạn?

f/ Căn đến đâu bạn?

g/ Căn đến đâu bạn?

h/ \(y=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)

Do \(0\le sin^22x\le1\Rightarrow\frac{1}{2}\le y\le1\)

\(y_{max}=1\) khi \(sin^22x=0\)

\(y_{min}=\frac{1}{2}\) khi \(sin^22x=1\)

t/ \(y=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)

\(y=1-3sin^2x.cos^2x=1-\frac{3}{4}\left(2sinx.cosx\right)^2\)

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

Tượng tự câu trên \(\Rightarrow\frac{1}{4}\le y\le1\)

\(y_{min}=\frac{1}{4}\) khi \(sin^22x=1\)

\(y_{max}=1\) khi \(sin^22x=0\)

Tốt nhất là bạn sử dụng công cụ gõ công thức

\(M=\sin^6x+\cos^6x\)(1)

Có công thức: \(y=\sin^{2n}x+\cos^{2n}x\)

\(\Rightarrow\max\limits_y=1;\min\limits_y=\dfrac{1}{2^{n-1}}\)

\(\Rightarrow\left(1\right)\) có \(\max\limits_M=1\)và \(\min\limits_M=\dfrac{1}{2^2}=\dfrac{1}{4}\)

\(y=\left(sin^2x+cos^2x\right)^2-3\left(sinx.cosx\right)^2\left(sin^2x+cos^2x\right)+2\)

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

\(0\le sin^22x\le1\Rightarrow\frac{9}{4}\le y\le3\)

\(y_{max}=3\) khi \(sin2x=0\Leftrightarrow x=\pm\frac{\pi}{2}\)

\(y_{min}=\frac{9}{4}\) khi \(sin^22x=1\Leftrightarrow x=\pm\frac{\pi}{4}\)

Biến đổi xong nó thành hàm nhìn gọn lắm :)

\(=\sin^6x+\cos^6x+\frac{3}{4}.\frac{1}{2}\left[\sin\left(2x+\frac{\pi}{4}-2x+\frac{\pi}{4}\right)+\sin\left(2x+\frac{\pi}{4}+2x-\frac{\pi}{4}\right)\right]\)

\(=1+\frac{3}{8}\left(\sin\frac{\pi}{4}+\sin4x\right)\)

Bạn biện luận nốt nhé

Bạn muốn bài cụ thể như nào? Giải tiếp hay giải thích kỹ các bước phân tích cho bạn?

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

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

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

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

Đáp án đúng : C

a/

\(y=2cos\left(3x+\frac{\pi}{12}\right).cos\left(-\frac{\pi}{4}\right)-4\)

\(=\sqrt{2}cos\left(3x+\frac{\pi}{12}\right)-4\)

Do \(-1\le cos\left(3x+\frac{\pi}{12}\right)\le1\Rightarrow-\sqrt{2}-4\le y\le\sqrt{2}-4\)

\(y_{max}=\sqrt{2}-4\) khi \(sin\left(3x+\frac{\pi}{12}\right)=1\)

\(y_{min}=-\sqrt{2}-4\) khi \(sin\left(3x+\frac{\pi}{12}\right)=-1\)

b/

\(y=2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)+2=2sin\left(x+\frac{\pi}{6}\right)+2\)

Do \(-1\le sin\left(x+\frac{\pi}{6}\right)\le1\)

\(\Rightarrow0\le y\le4\)

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 \(sin2x=1\)

e/

Đề câu này chắc chắn đúng chứ bạn?

f/

\(sin^4x+cos^4x=\frac{3}{4}\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\frac{3}{4}\)

\(\Leftrightarrow1-\frac{1}{2}\left(2sinx.cosx\right)^2=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{1}{2}sin^22x=0\)

\(\Leftrightarrow1-2sin^22x=0\)

\(\Leftrightarrow cos4x=0\)

\(\Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{4}\)

`y=sin^2 5x+cos^2 5x+3sin 2x-2`       `TXĐ: R`

`y=1+3sin 2x-2`

`y=3sin 2x-1`

Ta có: `-1 <= sin 2x <= 1`

`<=>-3 <= 3sin 2x <= 3`

`<=>-4 <= y <= 2`

    `=> y_[mi n]=4<=>sin 2x =-1<=>x=-\pi/4 + k\pi`  `(k in ZZ)`

         `y_[max] = 2<=>sin 2x=1<=>x=\pi/4+k\pi`  `(k in ZZ)`