8.

\(y=\left(cosx+1\right)^2-1\ge-1\Rightarrow y_{min}=-1\)

\(y=\left(cosx-1\right)\left(cosx+3\right)+3\le3\Rightarrow y_{max}=3\)

10.

\(y=2-\left(cosx+1\right)^2\le2\Rightarrow y_{max}=2\)

14.

Hàm tuần hoàn với chu kì \(T=\pi\)

15.

Đáp án A đúng

20.

\(-1\le sin\left(\frac{x}{2}+\frac{\pi}{7}\right)\le1\Rightarrow-5\le y\le-1\)

\(y_{max}=-1\) ; \(y_{min}=-5\)

2.

\(0\le\left|sinx\right|\le1\Rightarrow1\le y\le3\)

Min và max lần lượt là 3 và 1

3.

\(cos\left(x-\frac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

8.

\(y=\frac{1}{2}+\frac{1}{2}cos2x+2cos2x=\frac{1}{2}+\frac{5}{2}cos2x\le\frac{1}{2}+\frac{5}{2}.1=3\)

15.

Nó đi qua vô số điểm nên ko có 4 đáp án để chọn thì ko ai có thể trả lời câu này cho bạn cả

18.

\(y=\frac{sinx+2cosx+1}{sinx+cosx+2}\Leftrightarrow y.sinx+y.cosx+2y=sinx+2cosx+1\)

\(\Leftrightarrow\left(y-1\right)sinx+\left(y-2\right)cosx=1-2y\)

\(\left(y-1\right)^2+\left(y-2\right)^2\ge\left(1-2y\right)^2\)

\(\Leftrightarrow2y^2+2y-4\le0\Rightarrow-2\le y\le1\)

\(\Rightarrow y_{max}=1\)

1.

ĐKXĐ: \(\frac{1+x}{1-x}\ge0\Leftrightarrow-1\le x< 1\)

2.

\(cosx-cos3x\ne0\)

\(\Leftrightarrow cos3x\ne cosx\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)

3.

a. \(0\le\left|sinx\right|\le1\Rightarrow1\le y\le3\)

\(y_{min}=1\) khi \(\left|sinx\right|=1\)

\(y_{max}=3\) khi \(sinx=0\)

b. \(y=cosx+cos\left(x-\frac{\pi}{3}\right)=2cos\left(x-\frac{\pi}{6}\right).cos\frac{\pi}{6}=\sqrt{3}cos\left(x-\frac{\pi}{6}\right)\)

\(-1\le cos\left(x-\frac{\pi}{6}\right)\le1\Rightarrow-\sqrt{3}\le y\le\sqrt{3}\)

c. \(y=cos^22x+2cos2x+1-1=\left(cos2x+1\right)^2-1\ge-1\)

\(y_{min}=-1\) khi \(cos2x=-1\)

\(cos2x\le1\Leftrightarrow\left\{{}\begin{matrix}cos^22x\le1\\2cos2x\le2\end{matrix}\right.\) \(\Rightarrow y\le3\)

\(y_{max}=3\) khi \(cos2x=1\)

d. \(5-2cos^2x.sin^2x=5-\frac{1}{2}\left(2sinx.cosx\right)^2=5-\frac{1}{2}sin^22x\)

\(0\le sin^22x\le1\Rightarrow\frac{9}{2}\le5-\frac{1}{2}sin^22x\le5\)

\(\Rightarrow\sqrt{\frac{9}{2}}\le y\le\sqrt{5}\)

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

\(y=sin^2x-6sinx+10\)

\(y=sin^2x-6sinx-7+17=\left(sinx+1\right)\left(sinx-7\right)+17\le17\)

\(y_{max}=17\) khi \(sinx=-1\)

\(y=sin^2x-6sinx+5+5=\left(1-sinx\right)\left(5-sinx\right)+5\ge5\)

\(y_{min}=5\) khi \(sinx=1\)

Bài 1:

1: \(y=\frac{\sin x+2\cdot cosx+1}{2\cdot\sin x+cosx+3}\)

=>\(2y\cdot\sin x+y\cdot cosx+3y=\sin x+2\cdot cosx+1\)

=>\(\left(2y-1\right)\cdot\sin x+cosx\cdot\left(y-2\right)=1-3y\)

Để phương trình có nghiệm thì \(\left(2y-1\right)^2+\left(y-2\right)^2>=\left(1-3y\right)^2\)

=>\(4y^2-4y+1+y^2-4y+4\ge9y^2-6y+1\)

=>\(5y^2-8y+5-9y^2+6y-1\ge0\)

=>\(-4y^2-2y+4\ge0\)

=>\(y^2+\frac12y-1\le0\)

=>\(y^2+2\cdot y\cdot\frac14+\frac{1}{16}-\frac{17}{16}\le0\)

=>\(\left(y+\frac14\right)^2\le\frac{17}{16}\)

=>\(-\frac{\sqrt{17}}{4}\le y+\frac14\le\frac{\sqrt{17}}{4}\)

=>\(\frac{-\sqrt{17}-1}{4}\le y\le\frac{\sqrt{17}-1}{4}\)

=>\(y_{\min}=\frac{-\sqrt{17}-1}{4}\) và \(y_{\max}=\frac{\sqrt{17}-1}{4}\)

2: \(y=2\cdot\sin^2x-3\cdot\sin x\cdot cosx+cos^2x\)

\(=2\cdot\frac{1-cos2x}{2}-3\cdot\frac12\cdot\sin2x+\frac{1+cos2x}{2}\)

\(=1-cos2x-\frac32\cdot\sin2x+\frac12+\frac12\cdot cos2x\)

\(=-\frac32\cdot\sin2x-\frac12\cdot cos2x+\frac32=-\frac12\left(3\cdot\sin2x+cos2x-3\right)\)

\(=-\frac{\sqrt{10}}{2}\left(\frac{3}{\sqrt{10}}\cdot\sin2x+\frac{1}{\sqrt{10}}\cdot cos2x-\frac{3}{\sqrt{10}}\right)\)

\(=-\frac{\sqrt{10}}{2}\cdot\left\lbrack\sin\left(2x+\alpha\right)-\frac{3}{\sqrt{10}}\right\rbrack\) , với \(cosa=\frac{3}{\sqrt{10}};\sin a=\frac{1}{\sqrt{10}}\)

\(=-\frac{\sqrt{10}}{2}\cdot\sin\left(2x+\alpha\right)+\frac32\)

Ta có: \(-1\le\sin\left(2x+a\right)\le1\)

=>\(-1\cdot\frac{-\sqrt{10}}{2}\ge\frac{-\sqrt{10}}{2}\sin\left(2x+a\right)\ge1\cdot\frac{-\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)\le\frac{\sqrt{10}}{2}\)

=>\(\frac{-\sqrt{10}}{2}+\frac32\le\frac{-\sqrt{10}}{2}\cdot\sin\left(2x+a\right)+\frac32\le\frac{\sqrt{10}}{2}+\frac32\)

=>\(y_{\min}=\frac{-\sqrt{10}+3}{2};y_{\max}=\frac{\sqrt{10}+3}{2}\)

\(y=1-2\sin^2x-\sin x+3=-2\sin^2x-\sin x+4\)

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

Xét hàm f(t) trên [-1;1]

\(f\left(-1\right)=-2+1+4=3\)

\(f\left(1\right)=-2-1+4=1\)

\(f\left(-\frac{1}{4}\right)=-2.\frac{1}{16}+\frac{1}{4}+4=\frac{33}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}y_{max}=\frac{33}{8};"="\Leftrightarrow\sin x=-\frac{1}{4}\Rightarrow x=...\\y_{min}=1;"="\Leftrightarrow\sin x=1\end{matrix}\right.\)

a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)

\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)

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

b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)

\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)

\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)

c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)

\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)

\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)