1.
a, \(sin2x-\sqrt{3}cos2x=-1\)
\(\Leftrightarrow\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{3\pi}{4}+k\pi\end{matrix}\right.\)
Do tổng các hệ số thứ 1,2,3 là 46 nên ta có:\(C_n^0+C_n^1+C_n^2=46\)
\(\Leftrightarrow1+\dfrac{n!}{1!\left(n-1\right)!}+\dfrac{n!}{2!\left(n-2\right)!}=46\)
\(\Leftrightarrow1+n+\dfrac{\left(n-1\right)n}{2}=46\)
\(\Leftrightarrow n^2+n-90=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=9\\n=-10\left(loai\right)\end{matrix}\right.\)
Khai triển biểu thức: \(\left(x+\dfrac{1}{x}\right)^9\)
Hạng tử thứ k+1 trong biểu thức trên
\(\left(x+\dfrac{1}{x}\right)^9=C_9^{k+1}+\left(x^2\right)^{10-k}.\left(\dfrac{1}{x}\right)^{k+1}\)
đến đây mình chịu rùi hjhj b nào làm được giúp b kia với
1.
b, \(cos^2\left(x+\dfrac{\pi}{3}\right)+cos^2\left(x+\dfrac{2\pi}{3}\right)=\dfrac{1}{2}\left(sinx+1\right)\)
\(\Leftrightarrow cos^2\left(x+\dfrac{\pi}{3}\right)+cos^2\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\left(sinx+1\right)\)
\(\Leftrightarrow2cos^2\left(x+\dfrac{\pi}{3}\right)-1+2cos^2\left(x-\dfrac{\pi}{3}\right)-1=sinx-1\)
\(\Leftrightarrow cos\left(2x+\dfrac{2\pi}{3}\right)+cos\left(2x-\dfrac{2\pi}{3}\right)=sinx-1\)
\(\Leftrightarrow2cos2x.cos\dfrac{2\pi}{3}=sinx-1\)
\(\Leftrightarrow-cos2x=sinx-1\)
\(\Leftrightarrow2sin^2x-sinx=0\)
\(\Leftrightarrow sinx\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
Đến đây dễ rồi, tự làm tiếp.
