1.
a, \(\sqrt{3}sin2x+2cos^2x=0\)
\(\Leftrightarrow\sqrt{3}sin2x+2cos^2x-1=-1\)
\(\Leftrightarrow\sqrt{3}sin2x+cos2x=-1\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x=-\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow2x-\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
1.
c, ĐK: \(x\ne k2\pi\)
\(\dfrac{2sin\left(x+\dfrac{\pi}{6}\right)-cos2x}{cosx-1}\)
\(\Leftrightarrow\sqrt{3}sinx+cosx-cos2x=cosx-1\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cos2x=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=-\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\left(l\right)\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{4\pi}{3}+k2\pi\)
2.
ĐK: \(n\ge2;n\in N\)
\(A_n^2+C_{n+1}^3=10\left(n-1\right)\)
\(\Leftrightarrow\dfrac{n!}{\left(n-2\right)!}+\dfrac{\left(n+1\right)!}{3!.\left(n-2\right)!}=10\left(n-1\right)\)
\(\Leftrightarrow n\left(n-1\right)+\dfrac{\left(n+1\right)n\left(n-1\right)}{6}=10\left(n-1\right)\)
\(\Leftrightarrow6n^2-6n+n^3-n=60n-60\)
\(\Leftrightarrow n^3+6n^2-67n+60=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=-12\left(l\right)\\n=5\left(tm\right)\\n=1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow n=5\)
3.
Số hạng tổng quát trong khai triển:
\(T_{k+1}=C_{12}^k.\left(x^3\right)^{12-k}.\left(\dfrac{2}{x}\right)^k\left(k\in Z;0\le k\le12\right)\)
\(=2^kC_{12}^k.x^{36-3k}.x^{-k}\)
\(=2^kC_{12}^k.x^{36-4k}\)
Số hạng không chứa x tương ứng với k thỏa mãn:
\(36-4k=0\Leftrightarrow k=9\)
\(\Rightarrow\) Số hạng không chứa x là \(2^9.C_{12}^9=112640\).
Số hạng chứa \(x^{12}\) tương ứng với k thỏa mãn:
\(36-4k=12\Leftrightarrow k=6\)
\(\Rightarrow\) Số hạng chứa \(x^{12}\) là \(2^6C_{12}^6.x^{12}=59136x^{12}\)
\(\Rightarrow\) Hệ số của \(x^{12}\) là \(59136\).
1.
b, \(sin5x+sinx+cos4x+1=0\)
\(\Leftrightarrow2sin3x.cos2x+2cos^22x=0\)
\(\Leftrightarrow2cos2x\left(sin3x+cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x+cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos\left(\dfrac{\pi}{2}-3x\right)+cos2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\2cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right).cos\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\\cos\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{4}-\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{4}-\dfrac{5x}{2}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
