Ta có: \(u_n>0\)

Mặt khác: 

\(u_n=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}< \dfrac{1}{n}+\dfrac{1}{n}+...+\dfrac{1}{n}\\ \Leftrightarrow u_n< n.\dfrac{1}{n}=1\)

\(0< u_n< 1\) nên \(u_n\) bị chặn

Xét tính đơn điệu dãy \(u_n\)

\(u_{n+1}=\dfrac{1}{\left(n+1\right)+1}+\dfrac{1}{\left(n+1\right)+2}+...+\dfrac{1}{\left(n+1\right)+\left(n+1-1\right)}+\dfrac{1}{\left(n+1\right)+\left(n+1\right)}\\ =\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}\)

\(u_{n+1}-u_n=\dfrac{1}{2n+1}+\dfrac{1}{2n+2}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}+\dfrac{1}{2\left(n+1\right)}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}-\dfrac{1}{2\left(n+1\right)}\\ =\dfrac{2\left(n+1\right)-\left(2n+1\right)}{2\left(2n+1\right)\left(n+1\right)}=\dfrac{1}{2\left(2n+1\right)\left(n+1\right)}>0\)

\(\Rightarrow u_{n+1}>u_n\)

Dãy \(u_n\) là dãy tăng.

Vậy \(u_n\) bị chặn và tăng nghiêm ngặt nên \(u_n\) hội tụ. đpcm

1: \(cos\left(2x+15^0\right)=\frac{\sqrt2}{2}\)

=>\(\left[\begin{array}{l}2x+15^0=45^0+k\cdot360^0\\ 2x+15^0=-45^0+k\cdot360^0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=30^0+k\cdot360^0\\ 2x=-60^0+k\cdot360^0\end{array}\right.\)

=>\(\left[\begin{array}{l}x=15^0+k\cdot180^0\\ x=-30^0+k\cdot180^0\end{array}\right.\)

2:

ĐKXĐ: \(2x<>\frac{\pi}{2}+k\pi\)

=>\(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)

\(\sqrt3\cdot\tan2x-3=0\)

=>\(\sqrt3\cdot\tan2x=3\)

=>\(\tan2x=\sqrt3\)

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

=>\(x=\frac{\pi}{6}+\frac{k\pi}{2}\) (nhận)

3: \(\sin\left(3x+45^0\right)=\frac{\sqrt3}{2}\)

=>\(\left[\begin{array}{l}3x+45^0=60^0+k\cdot360^0\\ 3x+45^0=180^0-60^0+k\cdot360^0=120^0+k\cdot360^0\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=15^0+k\cdot360^0\\ 3x=75^0+k\cdot360^0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5^0+k\cdot120^0\\ x=25^0+k\cdot120^0\end{array}\right.\)

\(sin3x=sinx\)

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

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

\(\sin3x-\sin x=0\)

\(\Leftrightarrow\sin3x=\sin x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=k2\pi\\4x=\pi+k2\pi\end{matrix}\right.\)

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

Nêu hiện tượng phải không bạn?

Đề là gì thế em