b: \(cosx+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(cos\left(2\cdot\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(1-2\cdot\sin^2\left(\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(-2\cdot\sin^2\left(\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-1=0\)

=>\(2\cdot\sin^2\left(\frac{x}{2}\right)-3\cdot\sin\left(\frac{x}{2}\right)+1=0\)

=>\(\left(2\cdot\sin\left(\frac{x}{2}\right)-1\right)\left(\sin\left(\frac{x}{2}\right)-1\right)=0\)

TH1: \(\sin\left(\frac{x}{2}\right)-1=0\)

=>\(\sin\left(\frac{x}{2}\right)=1\)

=>\(\frac{x}{2}=\frac{\pi}{2}+k2\pi\)

=>\(x=\pi+k4\pi\)

TH2: \(2\cdot\sin\left(\frac{x}{2}\right)-1=0\)

=>\(\sin\left(\frac{x}{2}\right)=\frac12\)

=>\(\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}+k2\pi\\ \frac{x}{2}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k4\pi\\ x=\frac53\pi+k4\pi\end{array}\right.\)

Hình chữ nhật???

Tròn chứ!

chị hơn em 6 tuổi

Hiểu như này:

\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{b}{1+b}=3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+b}\right)\le3-\dfrac{9}{1+a+1+b+1+b}=\dfrac{3\left(a+2b\right)}{3+a+2b}\)

ghê quá :<

\(5x+\sqrt{9x^2-6x+1}\)

\(=5x+1-3x\)

=2x+1

C

C

3.

\(4sinx+cosx+2cos\left(x+\dfrac{\pi}{3}\right)=2\)

\(\Leftrightarrow4sinx+cosx+cosx-\sqrt{3}sinx=2\)

\(\Leftrightarrow\left(4-\sqrt{3}\right)sinx+2cosx=2\)

\(\Leftrightarrow\sqrt{23-4\sqrt{3}}\left(\dfrac{4-\sqrt{3}}{\sqrt{23-4\sqrt{3}}}sinx+\dfrac{2}{\sqrt{23-4\sqrt{3}}}cosx\right)=2\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}\right)=\dfrac{2}{\sqrt{23-4\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}=\pm arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{2}{\sqrt{23-4\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

4.

\(sinx+2cos\left(x+\dfrac{\pi}{3}\right)+4sin\left(x+\dfrac{\pi}{6}\right)+cosx=4\)

\(\Leftrightarrow sinx+cosx-\sqrt{3}sinx+2\sqrt{3}sinx+2cosx+cosx=4\)

\(\Leftrightarrow\left(1+\sqrt{3}\right)sinx+4cosx=4\)

\(\Leftrightarrow\sqrt{20+2\sqrt{3}}\left(\dfrac{1+\sqrt{3}}{\sqrt{20+2\sqrt{3}}}sinx+\dfrac{4}{\sqrt{20+2\sqrt{3}}}cosx\right)=4\)

\(\Leftrightarrow cos\left(x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}\right)=\dfrac{4}{\sqrt{20+2\sqrt{3}}}\)

\(\Leftrightarrow x-arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}=\pm arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2arccos\dfrac{4}{\sqrt{20+2\sqrt{3}}}+k2\pi\\x=k2\pi\end{matrix}\right.\)

S=ab

có trong sách nha em

\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)