Đề bài là: Tính cos2x 

Cảm ơn mn nhiều ạ!

`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`

`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`

`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`

`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`

`<=> cos2x - cos4x - cos2x = 1/2`

`<=> cos4x = cos(2π)/3`

`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)

\(sin3x.sinx+sin\left(x-\dfrac{\pi}{3}\right)cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin\left(2x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin\left(-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}cos2x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos4x+\dfrac{1}{2}=0\)

\(\Leftrightarrow2cos^22x-1+\dfrac{1}{2}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

\(\Rightarrow cos2x=\pm\dfrac{1}{2}\)

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8

c) Ta có: \(\sqrt{\sqrt{x}+3}=3\)

\(\Leftrightarrow\sqrt{x}+3=9\)

\(\Leftrightarrow\sqrt{x}=6\)

hay x=36

Ta có: \(\sqrt{x-2\sqrt{x-1}}=2\)

\(\Leftrightarrow x-2\sqrt{x-1}-4=0\)

\(\Leftrightarrow x-1-2\cdot\sqrt{x-1}\cdot1+1=4\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=4\)

\(\Leftrightarrow\sqrt{x-1}-1=2\)

\(\Leftrightarrow\sqrt{x-1}=3\)

\(\Leftrightarrow x-1=9\)

hay x=10

Hình mờ lắm ak bn!

Đặt \(\dfrac{x}{4}=\dfrac{y}{7}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)

Ta có: xy=112

\(\Leftrightarrow28k^2=112\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\cdot2=8\\y=7k=7\cdot2=14\end{matrix}\right.\)

Trường hợp 2: x=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-8\\y=7k=-14\end{matrix}\right.\)

bạn gửi hình lên đây kiểu j thế

cứu em với huhu