Đại số lớp 7

TC

a) \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}\)\(-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
b)\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
c) \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+.....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+......+\dfrac{1}{255.257}\)

DH
17 tháng 6 2017 lúc 17:06

a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)

câu b và c xem lại đề nha

Chúc bạn học tốt!!!

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DH
17 tháng 6 2017 lúc 17:21

b, \(\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+......+\dfrac{1}{3.2}+\dfrac{1}{2.1}\)

\(=\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}+.........+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\))

\(=\dfrac{1}{99}+\dfrac{1}{99}-1=\dfrac{2}{99}-1=\dfrac{-97}{99}\)

Chúc bạn học tốt!!!

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AT
17 tháng 6 2017 lúc 17:49

c/ \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1}{155\cdot257}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}+\dfrac{2}{255\cdot257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}+...+\dfrac{1}{255}-\dfrac{1}{257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{257}\right)=\dfrac{1}{2}\cdot\dfrac{256}{257}=\dfrac{128}{257}\)

p/s: thấy bn kia hơn 20' r` k lm tiếp nên t lm lun nhé!

Bình luận (4)
MS
17 tháng 6 2017 lúc 18:50

\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)\(=0+0+0+0+0+0+\dfrac{13}{15}\)

\(=\dfrac{13}{15}\)

\(\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-.........-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{96}-.....-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)

\(=0+1=1\)

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