Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(A=\frac{1}{2}-\frac{1}{17}\)

\(A=\frac{15}{34}\)

= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)

Ta có : 

\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2009.2012}+\frac{3}{2012.2015}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2009}-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2015}\)

\(A=\frac{1}{5}-\frac{1}{2015}\)

\(A=\frac{402}{2015}\)

Vậy \(A=\frac{402}{2015}\)

Chúc bạn học tốt ~ 

A =  402/2015 

\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2012.1015}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\)

\(A=\frac{1}{5}-\frac{1}{2015}\)

\(A=\frac{402}{2015}\)

\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{21.24}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{24}\)

\(=\frac{1}{5}-\frac{1}{24}\)

\(=\frac{19}{120}\)

\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)

=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)

= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)

= ​\(3\left(\frac{1}{2}-\frac{1}{17}\right)\)​​

=\(\frac{45}{34}\)

\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)​

=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)

=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)

=3(1/2-1/17)

=45/34

cô Nhung ơi k đúng cho con đi cô pls

\(H=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{197.200}\right)=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{197}-\frac{1}{200}\right)=3\cdot\left(\frac{1}{2}-\frac{1}{200}\right)==\frac{297}{200}\)

297/200 nha  kick nha

= 297/200

Dat A=3/5.8 +3/8.11 +.........+3/32.35

A=1/5-1/8+1/8-1/11+1/11-1/14+.............+1/32-1/35

A=1/5-1/35

A=6/35

=>x+6/35=-29/35

=>x=-29/35-6/35

=>x=-1

Câu hỏi của Nguyễn Ánh Ngân - Toán lớp 6 - Học toán với OnlineMath

Bạn tham khảo

\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\cdot\left(x+3\right)}=\frac{101}{1504}\)

\(\Rightarrow\frac{1}{3}\cdot\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\cdot\left(x+3\right)}\right)=\frac{101}{1504}\)

\(\Rightarrow\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1504}\)

\(\Rightarrow\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1504}\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}:\frac{1}{3}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1504}\cdot\frac{3}{1}=\frac{303}{1504}\)

- Đến đây tự tính nhé :v

#Amano Ichigo:

- Bn ơi kq của biểu thức tìm x phải bằng \(\frac{101}{1540}\) mới đúng nhé =)

Học tốt !! :v

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)

\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)

\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)

\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

<=>\(\frac{1}{x+3}=\frac{1}{308}\)

=>x+3=308

<=>x=305 (nhận)

Vậy x=305