Câu 5: \(2\cdot\sin3x-\sqrt3\cdot cosx=\sin x\)

=>\(\sin x+\sqrt3\cdot cosx=2\cdot\sin3x\)

=>\(\sin x\cdot\frac12+cosx\cdot\frac{\sqrt3}{2}=\sin3x\)

=>\(\sin\left(x+\frac{\pi}{3}\right)=\sin3x\)

=>\(\left[\begin{array}{l}3x=x+\frac{\pi}{3}+k2\pi\\ 3x=\pi-x-\frac{\pi}{3}+k2\pi=-x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 4x=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi=\frac{\pi}{6}+\frac{k2\pi}{2}\\ x=\frac16\pi+\frac{k\pi}{2}=\frac{\pi}{6}+\frac{k2\pi}{4}\end{array}\right.\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

=>Có 4 điểm biểu diễn

=>Chọn D

câu 4: C

Câu 5: \(2\cdot\sin3x-\sqrt3\cdot cosx=\sin x\)

=>\(\sin x+\sqrt3\cdot cosx=2\cdot\sin3x\)

=>\(\sin x\cdot\frac12+cosx\cdot\frac{\sqrt3}{2}=\sin3x\)

=>\(\sin\left(x+\frac{\pi}{3}\right)=\sin3x\)

=>\(\left[\begin{array}{l}3x=x+\frac{\pi}{3}+k2\pi\\ 3x=\pi-x-\frac{\pi}{3}+k2\pi=-x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 4x=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi=\frac{\pi}{6}+\frac{k2\pi}{2}\\ x=\frac16\pi+\frac{k\pi}{2}=\frac{\pi}{6}+\frac{k2\pi}{4}\end{array}\right.\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

=>Có 4 điểm biểu diễn

=>Chọn D

câu 4: C

\(2\cdot\sin x\left(cosx+1\right)=\sqrt3\cdot cos2x\)

=>\(\sin2x+2\cdot\sin x=\sqrt3\cdot cos2x\)

=>\(2\cdot\sin x=-\sin2x+\sqrt3\cdot cos2x\)

=>\(\sin x=-\frac12\cdot\sin2x+\frac{\sqrt3}{2}\cdot cos2x=cos2x\cdot\sin\left(\frac{\pi}{3}\right)-\sin2x\cdot cos\left(\frac{\pi}{3}\right)\)

=>\(\sin x=\sin\left(\frac{\pi}{3}-2x\right)\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{3}-2x+k2\pi\\ x=\pi-\frac{\pi}{3}+2x+k2\pi=\frac23\pi+2x+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=\frac{\pi}{3}+k2\pi\\ -x=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{9}+\frac{k2\pi}{3}\\ x=-\frac23\pi-k2\pi=-\frac23\pi+k2\pi\end{array}\right.\)

=>a=-2/3; b=1/9

a+b=-2/3+1/9=-6/9+1/9=-5/9

=>Chọn A

Câu 2: B

Câu 3: C

C

Chọn C

Đăng tách ra.

Câu 1: Ý C

PT \(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\) mà\(x\in\left(0;2\pi\right)\)

Có 3 nghiệm

Câu 2: Ý A

PT \(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\) mà \(0\le x< \dfrac{\pi}{2}\)

\(\Rightarrow x=\dfrac{\pi}{6}\)

tự luận thì bạn up từng câu 1 lên nhé =)) Đăng nguyên cái đề lên k có ai làm đâu =))))))

luc keo la F=P/v=75000(N)

trong luong cua tau la p=F/0,005 =15000000(N)

khoi luong m=p/10=15000000/10=1500000(kg)

Neu dung nho tick nha. Chu y phan biet P la cong suat

p la trong luong