Question

Bài 1 :TÍNH

A= \(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)+\(\dfrac{3}{11.14}\)+ . . . + \(\dfrac{3}{2006.2009}\)

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Answer 1

\(A=\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{2006.2009}\\ A=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{2006}-\dfrac{1}{2009}\\ A=\dfrac{1}{5}-\dfrac{1}{2009}=\dfrac{2004}{10045}\)

Answer 2

\(A=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{2006}-\dfrac{1}{2009}\)

\(=\dfrac{1}{5}-\dfrac{1}{2009}=\dfrac{2004}{10045}.\)

Answer 3

\(A=\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+.....+\dfrac{3}{2006.2009}\)

\(A=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+.....+\dfrac{1}{2006}-\dfrac{1}{2009}\)

\(A=\dfrac{1}{5}-\dfrac{1}{2009}\)

\(A=\dfrac{2004}{10045}\)