a,b,c,d tôi có giải ở câu hỏi trc rồi nha

\(e,tan\left(2x+\dfrac{\Pi}{3}\right)=\sqrt{3}\Rightarrow tan\left(2x+\dfrac{\Pi}{3}\right)=tan\dfrac{\Pi}{3}\Rightarrow2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{3}+k\Pi\Rightarrow2x=k\Pi\Rightarrow x=\dfrac{k\Pi}{2}\)\(f,2sin4x+\sqrt{3}=0\Rightarrow sin4x=-\dfrac{\sqrt{3}}{2}\Rightarrow sin4x=sin\left(\dfrac{-\Pi}{3}\right)\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{\Pi}{3}+k2\Pi\\4x=\Pi-\left(-\dfrac{\Pi}{3}\right)+k2\Pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-\Pi}{12}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{3}+\dfrac{k\Pi}{2}\end{matrix}\right.\)
\(h,sin\left(2x+\dfrac{\Pi}{4}\right)=sin\left(x-\dfrac{\Pi}{3}\right)\Rightarrow\left[{}\begin{matrix}2x+\dfrac{\Pi}{4}=x-\dfrac{\Pi}{3}+k2\Pi\\2x+\dfrac{\Pi}{4}=\Pi-x+\dfrac{\Pi}{3}+k2\Pi\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7\Pi}{12}+k2\Pi\\3x=\dfrac{13\Pi}{12}+k2\Pi\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7\Pi}{12}+k2\Pi\\x=\dfrac{13\Pi}{36}+\dfrac{k2\Pi}{3}\end{matrix}\right.\)
\(j,cot\left(3x-\dfrac{2\Pi}{3}\right)=\dfrac{-1}{\sqrt{3}}\Rightarrow cot\left(3x-\dfrac{2\Pi}{3}\right)=cot\left(-\dfrac{\Pi}{3}\right)\Rightarrow3x-\dfrac{2\Pi}{3}=\dfrac{-\Pi}{3}+k\Pi\Rightarrow3x=\dfrac{\Pi}{3}+k\Pi\)\(\Rightarrow x=\dfrac{\Pi}{9}+\dfrac{k\Pi}{3}\)

Quãng đường MN là: \(4.2.60=480\left(m\right)\)
Quãng đường NP là: \(2.3.60=360\left(m\right)\)
Vì vật chuyển động thẳng trên đường thẳng MNP và không đổi chiều chuyển động nên tốc độ trung bình bằng vận tốc trung bình: \(\dfrac{480+360}{120+180}=2,8\) (m/s)

1.\(sin8x=-\dfrac{1}{2}\Rightarrow sin8x=sin\dfrac{-\Pi}{6}\Rightarrow\left[{}\begin{matrix}8x=\dfrac{-\Pi}{6}+k2\Pi\\8x=\Pi-\left(-\dfrac{\Pi}{6}\right)+k2\Pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-\Pi}{48}+\dfrac{k\Pi}{4}\\x=\dfrac{7\Pi}{48}+\dfrac{k\Pi}{4}\end{matrix}\right.\)
2. \(sin9x+cos9x=0\Rightarrow sin9x=-cos9x\Rightarrow\dfrac{sin9x}{cos9x}=\dfrac{-cos9x}{cos9x}\Rightarrow tan9x=-1\)
\(\Rightarrow9x=\dfrac{-\Pi}{4}+k\Pi\Rightarrow x=\dfrac{-\Pi}{36}+\dfrac{k\Pi}{9}\)
3. \(cos5x=sin2x\Rightarrow cos5x=cos\left(\dfrac{\Pi}{2}-2x\right)\Rightarrow\left[{}\begin{matrix}5x=\dfrac{\Pi}{2}-2x+k2\Pi\\5x=-\dfrac{\Pi}{2}+2x+k2\Pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}7x=\dfrac{\Pi}{2}+k2\Pi\\3x=\dfrac{-\Pi}{2}+k2\Pi\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{14}+\dfrac{k2\Pi}{7}\\x=\dfrac{-\Pi}{6}+\dfrac{k2\Pi}{3}\end{matrix}\right.\)
 

Đáp án: pride
take pride in sth/sb = [be] proud of sth/sb: tự hào, hãnh diện về cái gì/ai

hahaha,tưởng là dc giải đặc biệt mới chia mà An ơi