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 

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

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

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

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

\(B=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2018.2021}\)

\(B=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2018}-\frac{1}{2021}\)

\(B=\frac{1}{5}-\frac{1}{2021}\)

\(B=\frac{2016}{10105}\)

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

297/200 nha  kick nha

= 297/200

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

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

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

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

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

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

=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

a/ Ta có: \(S=1+\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{27}-\frac{1}{30}\right)\)

\(S=1+\left(\frac{1}{2}-\frac{1}{30}\right)\)

\(S=1+\frac{7}{15}\)

\(S=\frac{22}{15}\)

b/ \(S=-4+\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{107}-\frac{1}{110}\right)\)

\(S=-4+\left(1-\frac{1}{110}\right)\)

\(S=-4+\frac{109}{110}\)

\(S=-3\frac{1}{110}\)

\(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)

\( B=\frac{1}{2}-\frac{1}{14}=\frac{3}{7}\)

\(C=\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)

\(C=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)

\(C=\frac{1}{1}-\frac{1}{16}=\frac{15}{16}\)

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