đáp án:307/1830

=1/3.3(1/2.5-1/5.8-1/8.11-...-1/302.305)

=1/3.(3/2.5-3/5.8-3/8.11-...-3/302.305)

=1/3(1/2-1/5-1/5-1/8-1/8-1/11-...-1/302-1/305)

=1/3[(1/2-1/305)+(1/5-1/5)+...+(1/302-1/302)

=1/3*(1/2-1/305)=1/3*(305/610-1/610)=1/3*304/610=152/915

hình như mình làm sai hoặc sai đề , sao số lớn ghê

Đề hình như bị sai ban ơi sửa lại

\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(A=\dfrac{1}{2}-\dfrac{1}{95}\)

\(A=\dfrac{93}{190}\)

\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)

\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)

\(3B=2.\dfrac{93}{190}\)

\(3B=\dfrac{93}{95}\)

\(\Rightarrow B=\dfrac{31}{95}\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)=\dfrac{4}{3}\cdot\dfrac{303}{610}=\dfrac{202}{305}\)

\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{302.305}\)

\(=4\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{302.305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{302.305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{305}\right)\)

\(=\dfrac{4}{3}.\dfrac{303}{610}\\ =\dfrac{202}{305}\)

=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

Chịu r

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

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

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

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

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

\(S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\)

\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\)

\(S=\frac{1}{2}-\frac{1}{101}\)

\(S=\frac{99}{202}\)

Bạn cho mình thêm 1/3 ở dòng 2 của bài làm mình nhé

Kết quả ra là : \(\frac{33}{202}\)

\(M= \dfrac{3^2}{2.5} +\dfrac{3^2}{5.8} +\dfrac{3^2}{8.11}+...+\dfrac{3^2}{98.101}\)

\(M= \) \( \dfrac{9}{2.5} +\dfrac{9}{5.8} +\dfrac{9}{8.11}+...+\dfrac{9}{98.101}\)

\(M=3(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+ \dfrac{3}{98.101})\)

\(M= 3(\dfrac{1}{2} -\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11})\)

\(M= 3(\dfrac{1}{2}-\dfrac{1}{11})\)

\(M=3(\dfrac{11}{22}- \dfrac{2}{22})\)

\(M=3.\dfrac{9}{22}\)

\(M=\dfrac{27}{22}\)