giúp mình nha 

Đặt A= \(\frac{1}{2}\)-\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)-\(\frac{1}{2^2}\)+....+\(\frac{1}{2^2}\)

=> 2A=1-\(\frac{1}{2}\)+\(\frac{1}{2^2}\)-\(\frac{1}{23}\)+...+\(\frac{1}{2^{98}}\)

=> 2A+A=1+\(\frac{1}{2^{99}}\)

=> 3A=1+\(\frac{1}{2^{99}}\)

=> A= \(\frac{1}{3}\)+\(\frac{1}{3.2^{99}}\)

Gọi a là USCLN của (n.(n+2) và n+1)

=> (n+1)⋮a (1) và n(n+2)⋮a (2)

=> n(n+1) ⋮ a (3)

Từ (2), (3), ta có

n(n+2)-n(n+1) ⋮ a

=> n⋮a (4)

Từ (1), (4), ta có:

(n+1)-n ⋮ a

=> 1⋮a

Vậy a=1

=> Biều thức \(\frac{n\left(n+2\right)}{n+1}\)tối giản với mọi n thuộc N (ĐPCM)

Gọi ƯCLN(n(n+2),n+1)=d

=>n+1(1):d và n(n+2):d(2)

=>n(n+1):d(3)

Từ (2) và (3)

=>n(n+2) - n(n+1):d

<=>n²+2n-n²-n:d hay n:d(4)

Từ (1) và (4)

=>(n+1)-n: d hay 1:d =>d=1

Vậy với mọi n thuộc N thì phân số n(n+2)/n tối giản.

1/2.3 - 8/3.5 + 4/5.9 - 22/9.13 + 31/13.18

=1/2.3- 4/3.5(2-1/3)-1/13.9 (22-31/2)

=\(\frac{1}{2.3}\)-\(\frac{4}{3.5}\).\(\frac{5}{3}\)-\(\frac{1}{13.9}\).\(\frac{13}{2}\)

=\(\frac{1}{6}\)-\(\frac{4}{9}\)-\(\frac{1}{18}\)

=\(\frac{3-8-1}{18}\)

=\(-\frac{1}{3}\)

\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+.......+\(\frac{1}{10^2}\)

\(\frac{1}{2^2}\)=\(\frac{1}{2.2}\)<\(\frac{1}{1.2}\)

\(\frac{1}{3^2}\)=\(\frac{1}{3.3}\)<\(\frac{1}{2.3}\)

\(\frac{1}{4^2}\)=\(\frac{1}{4.4}\)<\(\frac{1}{3.4}\)

\(\frac{1}{10^2}\)=\(\frac{1}{10.10}\)<\(\frac{1}{9.10}\)

=> S=\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+.....+\(\frac{1}{10.10}\)

=>S=\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+....+\(\frac{1}{10.10}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+....+\(\frac{1}{9.10}\)

=>S<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+.....+\(\frac{1}{9.10}\)

=>S<\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{9}\)-\(\frac{1}{10}\)

=>S<\(\frac{1}{1}\)-\(\frac{1}{10}\)=1-\(\frac{1}{10}\)

=>S<1-\(\frac{1}{10}\)<1

=>S<1

Vì \(\left|2x-1\right|\le4\) mà \(\left|2x-1\right|\ge0\) 

\(\Rightarrow2x-1\in\left\{0;1;2;3;4\right\}\)

Mà \(2x-1\)là số lẻ\(\Rightarrow2x-1\in\left\{1;3\right\}\)

\(\Rightarrow2x\in\left\{2;4\right\}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

Vậy \(x\in\left\{1;2\right\}\)