Ta có:\(A=\dfrac{1}{2}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{11}+...+\dfrac{31}{92}-\dfrac{32}{95}+\dfrac{32}{95}-\dfrac{33}{98}\)

                \(=\dfrac{1}{2}+\dfrac{33}{98}=\dfrac{82}{98}=\dfrac{41}{49}\)

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{95.98}\)

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

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

=> 3A = \(\frac{1}{2}-\frac{1}{98}\)

=> 3A = \(\frac{24}{49}\)

=> A = \(\frac{8}{49}\)

\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\)

\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\)

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

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)=\frac{1}{3}\cdot\frac{24}{49}=\frac{8}{49}\)

Sửa 95.98 thành 1/(95.98) nhá

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)

\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

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

\(A=\frac{1}{3}.\frac{24}{49}\)

\(A=\frac{8}{49}\)

Vậy ...........

A = 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/92.95 + 1/95.98

A = 1/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98 )

A = 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98 )

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

A = 1/3 . 24/49

A = 8/49

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\(A=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\frac{2}{8\cdot11}+...+\frac{2}{92\cdot95}+\frac{2}{95\cdot98}\)

\(A=\frac{2}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right]\)

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

\(A=\frac{2}{3}\left[\frac{1}{2}-\frac{1}{98}\right]=\frac{2}{3}\left[\frac{49}{98}-\frac{1}{98}\right]=\frac{2}{3}\cdot\frac{48}{98}=\frac{2}{3}\cdot\frac{24}{49}=\frac{2}{1}\cdot\frac{8}{49}=\frac{16}{49}\)

\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{92.95}+\frac{2}{95.98}\)

\(=\frac{2}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)

\(=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)

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

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

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

Sẽ không tính được giá trị cụ thể bạn nhé. Bạn xem lại. 

9992 + 185 =122222

94/95 nhé bạn