Chương III : Phân số

TA

\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.8}\)+\(\dfrac{2}{8.11}\)+...+\(\dfrac{2}{29.32}\)

ND
28 tháng 4 2017 lúc 20:12

\(\dfrac{2}{3.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{29.32}\)

= \(\dfrac{2}{3.5}+\left(\dfrac{2}{5.8}+\dfrac{2}{8.11}+\dfrac{2}{11.14}+...+\dfrac{2}{29.32}\right)\) =\(\dfrac{2}{15}+\dfrac{2}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{29.32}\right)\) = \(\dfrac{2}{15}+\dfrac{2}{3}\left(\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+\dfrac{14-11}{11.14}+...+\dfrac{32-29}{29.32}\right)\) =\(\dfrac{2}{15}+\dfrac{2}{3}\left(\dfrac{8}{5.8}-\dfrac{5}{5.8}+\dfrac{11}{8.11}-\dfrac{8}{8.11}+\dfrac{14}{11.14}-\dfrac{11}{11.14}+...+\dfrac{32}{29.32}-\dfrac{29}{29.32}\right)\) =\(\dfrac{2}{3}.\dfrac{1}{5}+\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{29}-\dfrac{1}{32}\right)\) =\(\dfrac{2}{3}\left(\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{32}\right)\)

= \(\dfrac{2}{3}\left(\dfrac{2}{5}-\dfrac{1}{32}\right)\)

=\(\dfrac{2}{3}\left(\dfrac{64}{160}-\dfrac{5}{160}\right)\)

=\(\dfrac{2}{3}.\dfrac{59}{160}\)

=\(\dfrac{59}{240}\)

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NT
28 tháng 4 2017 lúc 12:43

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{29}-\dfrac{1}{32}\\ =\dfrac{1}{3}-\dfrac{1}{32}\\ =\dfrac{32}{96}-\dfrac{3}{96}\\ =\dfrac{29}{96}\)

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NL
28 tháng 4 2017 lúc 12:45

\(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{29.32}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{8}+...+\dfrac{2}{29}-\dfrac{2}{32}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{3}-\dfrac{2}{32}\right)=\dfrac{1}{2}.\dfrac{29}{48}=\dfrac{29}{96}\)

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