\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{17\cdot19}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

\(A=\frac{1}{3}-\frac{1}{19}\)

\(A=\frac{16}{57}\)

Dấu "." là dấu nhân nhá ^^

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{17\cdot19}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{17}-\frac{1}{19}\)

\(=\frac{1}{3}-\frac{1}{19}\)

\(=\frac{16}{57}\)

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

\(=\frac{1}{3}-\frac{1}{19}\)

\(=\frac{16}{57}\)

Code : Breacker

\(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{17\times19}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

=\(\frac{1}{3}-\frac{1}{19}=\frac{19}{57}-\frac{3}{57}=\frac{19-3}{57}=\frac{16}{57}\)

=（2-1）*（2+1）+（4-1）*（4+1）+ ...+(2n-1)*(2n+1) =(2^2-1)+(4^2-1)+...+(4n^2-1) =(2^2+4^2+...+4n^2)-(1+1+...+1) =4(1^2+2^2+...n^2)-n n(n+1)(2n+1)/6： 1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6n^2=n 1x3+3x5+5x7+7x9+...+17x19 =4(1^2+2^2+...n^2)-n =4*n(n+1)(2n+1)/6-n； n=10，1x3+3x5+5x7+7x9+...+17x19=1530

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Leftrightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}\right)=2.\frac{8}{17}\)

\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{16}{17}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)

\(\Rightarrow x+2=17\Rightarrow x=15\)

x là số lẻ vậy x có thể là: 1 ; 3 ; 5 ; 7 ; 9

  Còn lại bạn tự giải nha! Cứ dùng phương pháp loại suy thử với từng số là ra! dễ mà

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{8}{17}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{8}{17}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{16}{17}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{17}\)

\(\Rightarrow x+2=17\Rightarrow x=15\)

a) Ta có: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}=1\text{-}\frac{1}{3}+\frac{1}{3}\text{-}\frac{1}{5}+...+\frac{1}{11}\text{-}\frac{1}{13}=1\text{-}\frac{1}{13}=\frac{12}{13}\)

Thay vào ta có:

\(\frac{12}{13}+x=\frac{24}{13}\Rightarrow x=\frac{24}{13}\text{-}\frac{12}{13}\Rightarrow x=\frac{12}{13}\)

\(\frac{2^2}{1x3}\)x \(\frac{4^2}{3x5}\)x \(\frac{6^2}{5x7}\) x \(\frac{8^2}{7x9}\)

= \(\frac{4}{3}\)x \(\frac{16}{15}\)x \(\frac{36}{35}\)x \(\frac{64}{63}\)

= \(1.486077098\)

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\right)x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{21}\right)\right]x=\frac{9}{7}\)

\(\left(\frac{1}{2}.\frac{2}{7}\right)x=\frac{9}{7}\)

\(\frac{1}{7}.x=\frac{9}{7}\)

\(x=\frac{9}{7}\div\frac{1}{7}\)

\(x=9\)

Vậy ...

ngày mai mik làm đc ko

ok ai giải được giúp mik nha chiều mai mik phải nộp rồi