Question

[học]

1/

\(Sm=\frac{m}{2}\left(2U_1+\left(m-1\right)d\right)\)

\(Sn=\frac{n}{2}=\left(2U_1+\left(n-1\right)d\right)\)

2/

a,\(3\tan\left(2x40^o\right)-\sqrt{3}=0\)

b,\(\cos4x-2\cos^23x+cos2x=0\)

Answer 1

[Hoàng Thị Thanh Huyền]

1/

\(Sm=\frac{m}{2}\left(2U_1+\left(m-1\right)d\right)\)

\(Sn=\frac{n}{2}\left(2U_1+\left(n-1\right)d\right)\)

\(\Rightarrow\frac{Sm}{Sn}=\frac{m\left[2U+_1\left(m-1\right)d\right]}{n\left[2U_1+\left(n-1\right)\right]}=\frac{m^2}{n^2}\)

\(\Rightarrow\frac{m}{n}=\frac{2U_1\left(m-1\right)d}{2U_1+\left(n-1\right)d}\)

\(\frac{Um}{Un}=\frac{U_1+\left(m-1\right)d}{U_1\left(n-1\right)d}\)

2/

a,\(3\tan\left(2x+40^o\right)\sqrt{3}=0\)

\(\Leftrightarrow tan\left(2x+40^o\right)=\frac{1}{\sqrt{3}}-tan30^o\)

\(\Rightarrow2x+40^o=30^o+k.180^o\)         \(\left(k\in Z\right)\)

\(\Leftrightarrow x=-5^o+k.90^o\)

b,\(\cos4x-2\cos^23x+\cos2x=0\)

\(\Leftrightarrow\left(\cos4x+\cos2x\right)-2cos^23x=0\)

\(\Leftrightarrow2cos\)\(3x\)\(cos\)\(x-2cos^23x=0\)

\(\Leftrightarrow\cos3x\left(\cos x-\cos3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\cos3x=0\\\cos x=\cos3x\end{cases}}\)

+\(\cos3x=0\Rightarrow3x=\frac{\pi}{2}+k\pi\left(k\inℤ\right)\)

\(\Leftrightarrow x=\frac{\pi}{6}+k\frac{\pi}{3}\)

+\(\cos x=\cos3x\Leftrightarrow\orbr{\begin{cases}3x=x+t2\pi\\3x=-3+t2\pi\end{cases}}\left(t\inℤ\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=t\pi\\x=\frac{t\pi}{2}\end{cases}}\Leftrightarrow x=\frac{t\pi}{2}\)

Vậy có No là \(x=\frac{\pi}{6}+k\frac{\pi}{3},x=\frac{t\pi}{2}\)