Đề là như thế này à bạn?

\(A^2_{2n}-C^3_{3n}=15n-4n^2\left(n\inℕ^∗\right)\)

\(\Rightarrow\frac{\left(2n\right)!}{\left(2n-2\right)!}-\frac{\left(3n\right)!}{3!\left(3n-3\right)}=15n-4n^2\)

\(\Rightarrow\left(2n-1\right)2n-\frac{\left(3n-2\right)\left(3n-1\right)3n}{6}=15n-4n^2\)

\(\Rightarrow6\left(4n^2-2n\right)-\left(9n^2-9n+2\right)3n=6\left(15n-4n^2\right)\)

\(\Rightarrow24n^2-12n-27n^3+27n^2-6n-90n+24n^2=0\)

\(\Rightarrow-27n^3+75n^2-108n=0\)(Vô nghiệm \(\forall n\inℕ^∗\))

ta có \(\hept{\begin{cases}\sqrt{2}\left(sinx+cosx\right)=2sin\left(x+\frac{\pi}{4}\right)\\sinx.cosx=\frac{1}{2}sin2x=-\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)=-\frac{1-2sin^2\left(x+\frac{\pi}{3}\right)}{2}\end{cases}}\)

Vậy phương trình \(\Leftrightarrow2sin\left(x+\frac{\pi}{4}\right)+\frac{1-2sin^2\left(x+\frac{\pi}{4}\right)}{2}=1\)

Đặt \(sin\left(x+\frac{\pi}{4}\right)=a\Rightarrow PT\Leftrightarrow2a+\frac{1-2a^2}{2}=1\Leftrightarrow\orbr{\begin{cases}a=1+\frac{1}{\sqrt{2}}\\a=1-\frac{1}{\sqrt{2}}\end{cases}}\)

vì sin <1 nên \(sin\left(x+\frac{\pi}{4}\right)=1-\frac{1}{\sqrt{2}}\)có 4 nghiệm trên \(\left(0,2\pi\right)\)

a) \(\frac{1}{2}\)(cos2x+cosx)cosx+\(\frac{1}{2}\)(cos2x-cosx)sinx=\(\frac{1}{2}\)

<=>cos2x.cosx+cos^x+cos2xsinx-cosxsinx=1

<=>cos2xcosx-sin²x+cos2xsinx-cosxsinx=0

<=>(sinx+cosx)(cos2x-sinx)=0

\(sin\left(x+\frac{\pi}{2}\right)\)\(=cosx\)\(=>sin^4\left(x+\frac{\pi}{2}\right)=cos^4x\)

\(VT=sin^4x-cos^4x=sin^2x-cos^2x=-cos2x\)

\(VP=4sin\frac{x}{2}cos\frac{x}{2}cosx=2sinxcosx=sin2x\)

\(=>sin2x=-cos2x\)

\(a,sin2x-2sinx+cosx-1=0\)

\(\Leftrightarrow2sinxcosx-2sinx+cosx-1=0\)

\(\Leftrightarrow2sinx\left(cosx-1\right)+cosx-1=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(2sinx+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=1\\sinx=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2k\pi\\x=\frac{-\pi}{6}+2k\pi\end{cases}}}\)

\(b,\sqrt{2}\left(sinx-2cosx\right)=2-sin2x\)

\(\Leftrightarrow\sqrt{2}sinx-2\sqrt{2}cosx-2+2sinxcosx=0\)

\(\Leftrightarrow\sqrt{2}sinx\left(1+\sqrt{2}cosx\right)-2.\left(\sqrt{2}cosx+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{2}cosx+1\right)\left(\sqrt{2}sinx-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{-\sqrt{2}}{2}\\sinx=\frac{2\sqrt{2}}{2}\left(l\right)\end{cases}}\)(vì \(-1\le sinx\le1\))

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3\pi}{4}+2k\pi\\x=\frac{5\pi}{4}+2k\pi\end{cases}}\)

\(c,\frac{1}{cosx}-\frac{1}{sinx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow\frac{sinx-cosx}{sinx.cosx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow\frac{-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)}{sinx.cosx}=2\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

\(\Leftrightarrow sin2x+1=0\)

\(\Leftrightarrow sin2x=-1\)

\(\Leftrightarrow2x=\frac{3\pi}{2}+2k\pi\)

\(\Leftrightarrow x=\frac{3\pi}{4}+k\pi\)

100 + 9900 = 10000

ta có 

\(-1\le cos\left(2x+\frac{\pi}{4}\right)\le1\Leftrightarrow-2\le2cos\left(2x+\frac{\pi}{4}\right)\le2\)

\(\Leftrightarrow5\le5+2cos\left(2x+\frac{\pi}{4}\right)\le7\Leftrightarrow\sqrt{5}\le\sqrt{5+2cos\left(2x+\frac{\pi}{4}\right)}\le\sqrt{7}\)

Vậy \(2-3\sqrt{5}\ge y\ge2-3\sqrt{7}\)

vậy \(\hept{\begin{cases}min\left(y\right)=2-3\sqrt{7}\\max\left(y\right)=2-3\sqrt{5}\end{cases}}\)

Ta thấy tâm vị tự \(I\left(1;-1\right)\) cũng là tâm của đường tròn \(\left(C\right)\). Do đó \(\left(C'\right),\left(C\right)\) đồng tâm

Suy ra tỉ số vị tự \(k=\frac{R'}{R}=\frac{IM}{R}=\frac{5}{4}\) thì \(\left(C'\right)\) đi qua M.