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

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

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

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

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

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

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

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

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

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

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

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

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

\(\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+...+\frac{2}{302\cdot305}\)

=\(\frac{2}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{302\cdot305}\right)\)

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

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

=\(\frac{2}{3}\cdot\frac{303}{610}\)

=\(\frac{101}{305}\)

\(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{14.17}=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{14.17}\right)\)

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

                                                                   \(=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{17}\right)=\frac{2}{3}.\frac{15}{34}=\frac{5}{17}\)

\(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{14.17}\)

\(=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\right)\)

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

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

\(=\frac{2}{3}.\left(\frac{17}{34}-\frac{2}{34}\right)\)

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

     \(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{14.17}\)

\(=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\right)\)

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

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

\(=\frac{2}{3}.\frac{16}{34}=\frac{16}{51}\)

A=2/2.5+2/5.8+2/8.11+...+2/95.98

=2/3.(3/2.5+3/5.8+3/8.11+...+3/95.98)

=2/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+1/95-1/98)

=2/3.(1/2-1/98)

=2/3.24/49

=16/49

VẬY A=16/49

\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\)

\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)

$A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}$

$A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}$

$A=\frac{1}{1}-\frac{1}{99}$

$A=\frac{98}{99}$

=

21​−51​+51​−71​+....+951​−981​

=12−198=21​−981​tự làm tiếp nha ( giống câu a)

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{302.305}\)

=\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)

=\(\frac{1}{5}-\frac{1}{305}\)

=\(\frac{12}{61}\)

tick cho mik nha

\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{95.98}\)

\(A=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)

\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)

\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)

\(A=\frac{2}{3}.\frac{24}{49}\)

\(A=\frac{16}{49}\)

\(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{95.98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{98}\)

\(\Leftrightarrow\frac{3}{2}A=\frac{48}{98}=\frac{24}{49}\)

\(\Leftrightarrow A=\frac{24}{49}\div\frac{3}{2}\)

\(\Leftrightarrow A=\frac{48}{147}\)

nguy thế,2 đs khác nhau