Bài 2: Cộng, trừ số hữu tỉ

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Kinomoto Sakura
13 tháng 7 2021 lúc 16:59

A=\(\dfrac{1}{2}\)\(\dfrac{2}{3}\)+\(\dfrac{3}{4}\)\(\dfrac{4}{5}\)+\(\dfrac{5}{6}\)\(\dfrac{6}{7}\)\(\dfrac{5}{6}\)+\(\dfrac{4}{5}\)\(\dfrac{3}{4}\)+\(\dfrac{2}{3}\)\(\dfrac{1}{2}\)

A=(\(\dfrac{1}{2}\)\(\dfrac{1}{2}\))+(−\(\dfrac{2}{3}\)+\(\dfrac{2}{3}\))+(\(\dfrac{3}{4}\)\(\dfrac{3}{4}\))+(−\(\dfrac{4}{5}\)+\(\dfrac{4}{5}\))+(\(\dfrac{5}{6}\)\(\dfrac{5}{6}\))−\(\dfrac{6}{7}\)

A=0+0+0+0+0−\(\dfrac{6}{7}\)

A=−\(\dfrac{6}{7}\)

Kinomoto Sakura
13 tháng 7 2021 lúc 17:13

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