ĐKXĐ: ..

\(\frac{sin3x+sinx+sin2x}{cos3x+cosx+cos2x}=\sqrt{3}\)

\(\Leftrightarrow\frac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\sqrt{3}\)

\(\Leftrightarrow\frac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=\sqrt{3}\)

\(\Leftrightarrow tan2x=\sqrt{3}\)

\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

`A=[sin x+sin 2x+sin 3x]/[cos x+cos 2x+cos 3x]`

`A=[(sin x+sin 3x)+sin 2x]/[(cos x+cos 3x)+cos 2x]`

`A=[2sin 2x.cos (-x)+sin 2x]/[2cos 2x.cos (-x)+cos 2x]`

`A=[sin 2x(2cos(-x)+1)]/[cos 2x(2cos(-x)+1)]`

`A=[sin 2x]/[cos 2x]=tan 2x`.

\(MSC\left(3;4;24\right)=24\)

⇒ \(\dfrac{1}{3}=\dfrac{1\times8}{3\times8}=\dfrac{8}{24}\)

\(\dfrac{1}{4}=\dfrac{1\times6}{4\times6}=\dfrac{6}{24}\)

\(\dfrac{1}{24}\) ( giữ nguyên phân số )

nhớ ghi MSC nữa nha

ui cj em giỏi thía

\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)

\(A=\frac{sin2x+sin6x+cos7x+cos3x}{sin3x-sinx}=\frac{2sin4x.cos2x+2cos5x.cos2x}{2cos2x.sinx}=\frac{2cos2x\left(sin4x+cos5x\right)}{2cos2x.sinx}\)

\(=\frac{sin4x+cos5x}{sinx}\)

ĐKXĐ: x>=0

\(Q=\dfrac{x-8}{\sqrt{x}+1}=\dfrac{x-1-7}{\sqrt{x}+1}\)

\(=\sqrt{x}-1-\dfrac{7}{\sqrt{x}+1}\)

=\(\sqrt{x}+1-\dfrac{7}{\sqrt{x}+1}-2\)

=>\(Q>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{7}{\sqrt{x}+1}}-2=2\sqrt{7}-2\)

Dấu '=' xảy ra khi \(\left(\sqrt{x}+1\right)^2=7\)

=>\(\sqrt{x}+1=\sqrt{7}\)

=>\(\sqrt{x}=\sqrt{7}-1\)

=>\(x=8-2\sqrt{7}\)

\(y^2=sin2x+cos2x+2\sqrt{sin2x.cos2x}\)

Đặt \(sin2x+cos2x=t\Rightarrow t\in\left[1;\dfrac{1+\sqrt{3}}{2}\right]\)

\(sin2x.cos2x=\dfrac{t^2-1}{2}\)

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

\(f'\left(t\right)=1+\dfrac{2t}{\sqrt{2\left(t^2-1\right)}}>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow y^2\le f\left(\dfrac{1+\sqrt{3}}{2}\right)=\dfrac{\left(1+\sqrt[4]{3}\right)^2}{2}\)

\(\Rightarrow y\le\dfrac{1+\sqrt[4]{3}}{\sqrt{2}}\)