1.

Chọn 3 chữ số còn lại từ 7 chữ số còn lại: \(C_7^3=35\) cách

Hoán vị 5 chữ số: \(5!=120\)

Số số thỏa mãn: \(35.120=4200\) số

2.

a.

\(\sqrt{3}sin5x-2sin5x.cos5x=0\)

\(\Leftrightarrow sin5x\left(\sqrt{3}-2cos5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\cos5x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{5}\\x=\pm\frac{\pi}{30}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=cos6x\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos6x\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=2x-\frac{\pi}{3}+k2\pi\\6x=\frac{\pi}{3}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{24}+\frac{k\pi}{4}\end{matrix}\right.\)

PT \(\Leftrightarrow \sin^2x+\cos^2x-2\sin x\cos x = -2(\cos^2 x - \sin^2 x)\)

\(\Leftrightarrow (\sin x-\cos x)^2 = 2(\sin x + \cos x)(\sin x - \cos x)\)

\(\Leftrightarrow (\sin x-\cos x)(\sin x - \cos x-2\sin x -2\cos x) = 0\)

\(\Leftrightarrow -(\sin x-\cos x)(\sin x + 3\cos x) = 0\)

Đến đây bạn giải tiếp nhé 

\(\Leftrightarrow2sin5x.sinx+1=2cos4x.sinx+2cos2x.sinx+3sinx\)

\(\Leftrightarrow2sin5x.sinx+1=sin5x-sin3x+sin3x-sinx+3sinx\)

\(\Leftrightarrow2sin5x.sinx-sin5x-2sinx+1=0\)

\(\Leftrightarrow sin5x\left(2sinx-1\right)-\left(2sinx-1\right)=0\)

Ta có: \(\left(2\cdot cos2x-1\right)\left(\sin2x+cos2x\right)=1\)

=>\(2\cdot cos2x\cdot\sin2x+2\cdot cos^22x-\sin2x-cos2x=1\)

=>\(\sin4x+1+cos4x-\sin2x-cos2x=1\)

=>\(\sin4x+cos4x-\sin2x-cos2x=0\)

=>\(\left(\sin4x-\sin2x\right)+\left(cos4x-cos2x\right)=0\)

=>\(2\cdot cos\left(\frac{4x+2x}{2}\right)\cdot\sin\left(\frac{4x-2x}{2}\right)+\left(-2\right)\cdot\sin\left(\frac{4x+2x}{2}\right)\cdot\sin\left(\frac{4x-2x}{2}\right)=0\)

=>\(2\cdot cos3x\cdot\sin x-2\cdot\sin3x\cdot\sin x=0\)

=>\(\sin x\cdot\left(cos3x-\sin3x\right)=0\)

TH1: sin x=0

=>\(x=\frac{\pi}{2}+k2\pi\)

TH2: cos3x-sin 3x=0

=>sin 3x-cos3x=0

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=0\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=0\)

=>\(3x-\frac{\pi}{4}=k\pi\)

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

=>\(x=\frac{\pi}{12}+\frac{k\pi}{3}\)

Giải phương trình hộ mình với

P/t \(\Leftrightarrow2cos2x.sin2x-sin2x+2cos^22x-cos2x-1=0\)

\(\Leftrightarrow sin4x-sin2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sinx.cos3x-2sin3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\cos3x=sin3x\left(2\right)\end{matrix}\right.\) 

(1) \(\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) \(\Leftrightarrow sin3x-cos3x=0\)  \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)

Vậy ... 

Với \(cosx=0\) ko phải nghiệm

Với \(cosx\ne0\)

\(\Rightarrow\left(2sin5x-1\right)\left(2cos2x.cosx-cosx\right)=2sinx.cosx\)

\(\Leftrightarrow\left(2sin5x-1\right)\left(cos3x+cosx-cosx\right)=sin2x\)

\(\Leftrightarrow cos3x\left(2sin5x-1\right)=sin2x\)

\(\Leftrightarrow2sin5x.cos3x-cos3x=sin2x\)

\(\Leftrightarrow sin8x+sin2x-cos3x=sin2x\)

\(\Leftrightarrow sin8x=cos3x=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow...\)

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