\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)

\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=3.\left(1-\frac{1}{10}\right)\)

\(A=3.\frac{9}{10}=\frac{27}{10}\)

\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)

\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

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

\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)

\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)

a) Lấy A chia 3

b) Lấy B nhân 3/2

15544525 nfhg

Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{32.35}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{32}-\frac{1}{35}\)

\(A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)

\(\Rightarrow x+\frac{6}{35}=-\frac{2}{7}\Rightarrow x=-\frac{2}{7}-\frac{6}{35}=-\frac{16}{35}\)

Chịu r

=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14

=1/2-1/14

=7/14-1/14=6/14=3/7

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 

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

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

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

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

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

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

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{101}{1540}\)

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+1}\right)=\frac{101}{1540}\)

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

=> x + 1 = 380 => x = 308 - 1 => x = 307

Vậy x = 307

=1/3(3/5.8+3/8.11+............+1/x(x+3)=101/1540

=.1/3(1/5.8+1/8.11+......1/x(x+3)=101/1540

=1/3(1/5-1/8+1/8-1/11+...........1/x-1/x+3=101/1540

=>1/3(1/5-1/x+3)=101/1540

=>1/5-1/x+3=101/1540 chia 1/3 =303/1540

=>1/x+3=   1/308

...........

dơn giản như đan rổ

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)

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

\(=\dfrac{1}{2}-\dfrac{1}{17}\)

\(=\dfrac{15}{34}\)

Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)

\(=\frac{1}{5}-\frac{1}{108}\)

\(=\frac{108}{540}-\frac{5}{540}=\frac{103}{540}\)

Ta có: 

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)

= \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)

= \(\frac{1}{5}-\frac{1}{108}\)

= \(\frac{103}{540}\)

#)Giải :

Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{105.108}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{105}-\frac{1}{108}\)

\(A=\frac{1}{5}-\frac{1}{108}\)

\(A=\frac{103}{540}\)