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\(n\ge2\)
\(\frac{3.\left(n+1\right)!}{3!.\left(n-2\right)!}-\frac{3.n!}{\left(n-2\right)!}=52\left(n-1\right)\)
\(\Leftrightarrow\frac{\left(n+1\right)n\left(n-1\right)}{2}-3n\left(n-1\right)=52\left(n-1\right)\)
\(\Leftrightarrow n\left(n+1\right)-6n=104\)
\(\Leftrightarrow n^2-5n-104=0\Rightarrow\left[{}\begin{matrix}n=13\\n=-8\left(l\right)\end{matrix}\right.\)
a/ ĐKXĐ: ...
\(y'=\frac{1}{2\sqrt{x+3}}-\frac{1}{2\sqrt{1-x}}\)
\(y'=0\Leftrightarrow\frac{1}{2\sqrt{x+3}}=\frac{1}{2\sqrt{1-x}}\Leftrightarrow\sqrt{x+3}=\sqrt{1-x}\)
\(\Leftrightarrow x+3=1-x\Rightarrow x=-1\)
b/ \(y'=\frac{sinx-x.cosx}{sin^2x}\)
c/ \(y'=\frac{cosx}{2\sqrt{x}}-\sqrt{x}.sinx\)
\(\left(sin\dfrac{x}{2}-cox\dfrac{x}{2}\right)^2+\sqrt{3}cosx=2sin5x+1\)
⇔\(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}cos\dfrac{x}{2}+\sqrt{3}cosx=2sin5x+1\)
⇔\(1-sinx+\sqrt{3}cosx=2sin5x+1\)
⇔\(sin\left(\dfrac{\Pi}{3}-x\right)=sin5x\)
\(2sinx\left(\sqrt{3}cosx+sinx+2sin3x\right)=1\)
⇔\(2\sqrt{3}sinxcosx+2sin^2x+4sinxsin3x=1\)
⇔\(\sqrt{3}sin2x+1-cos2x+cos2x-2cos4x=1\)
⇔\(\sqrt{3}sin2x+cos2x=2cos4x\)
⇔\(cos\left(2x-\dfrac{\Pi}{3}\right)=cos4x\)
Lâu lắm ko động tới đạo hàm hay giải phương trình lượng giác nên nhớ sương sương thôi, có gì thông cảm với ạ :<
\(y'=\frac{2.\left(-17\right).\sin17x}{17}-\frac{\sqrt{3}.5.\cos5x+5.\sin5x}{5}+2\)
\(y'=2\Leftrightarrow-2.\sin17x-\left(\sqrt{3}.\cos5x+\sin5x\right)=0\)
\(\Leftrightarrow-2.\sin17x-2\left(\frac{\sqrt{3}}{2}.\cos5x+\frac{1}{2}\sin5x\right)=0\)
\(\Leftrightarrow\sin17x+\sin\left(\frac{\pi}{6}+5x\right)=0\)
\(\Leftrightarrow\sin17x=-\sin\left(\frac{\pi}{6}+5x\right)\)
\(\Leftrightarrow\sin\left(-17x\right)=\sin\left(\frac{\pi}{6}+5x\right)\)
Đến đây chắc bạn giải được rồi :v
\(sin3x=cos\left(x+3\right)\)
\(\Leftrightarrow cos\left(x+3\right)=cos\left(\frac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=\frac{\pi}{2}-3x+k2\pi\\x+3=3x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{4}+\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{3}{2}+\frac{\pi}{4}+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow sin2x\left(1+\sqrt{2}sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sinx=-\frac{\sqrt{2}}{2}=sin\left(-\frac{\pi}{4}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow2sin2x.cos2x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x-\frac{1}{2}sin4x+\frac{1}{2}sinx=0\)
\(\Leftrightarrow sin4x=-sinx=sin\left(-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=-x+k2\pi\\4x=\pi+x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{5}\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
e/
\(sin\left(\frac{3\pi}{2}-sinx\right)=1\)
\(\Leftrightarrow\frac{3\pi}{2}-sinx=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sinx=\pi+k2\pi\)
Mà \(-1\le sinx\le1\Rightarrow-1\le\pi+k2\pi\le1\)
\(\Rightarrow\) Không tồn tại k nguyên thỏa mãn
Pt đã cho vô nghiệm
f/
\(cos^2x-sin^2x+sin4x=0\)
\(\Leftrightarrow cos2x+2sin2x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
\(\sqrt{3}.cos3x-sin3x=2\) \(\Leftrightarrow\dfrac{\sqrt{3}}{2}.cos3x-\dfrac{1}{2}.sin3x=1\)\(\Leftrightarrow sin\dfrac{\pi}{3}.cos3x-cos\dfrac{\pi}{3}.sin3x=1\)
\(\Leftrightarrow sin\left(\dfrac{\pi}{3}-3x\right)=1\)\(\Leftrightarrow\dfrac{\pi}{3}-3x=\dfrac{\pi}{2}+k2\pi\)\(\Leftrightarrow3x=\dfrac{\pi}{3}-\dfrac{\pi}{3}-k2\pi\)\(\Leftrightarrow x=-\dfrac{\pi}{6}-\dfrac{2k\pi}{3}\).