\(a,A=\frac{1}{2}-\frac{1}{3}=\frac{1}{6};B=\frac{1}{3}-\frac{1}{4}=\frac{1}{12};C=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\)\(b,A+B=\frac{1}{6}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4};A+B+C=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{10+5+3}{60}=\frac{18}{60}=\frac{3}{10}\)\(c,D=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)\(E=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)\(=-\left(-\frac{1}{99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+\frac{1}{97\cdot96}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)\(=-\left(-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)\(=-\left(-\frac{1}{99}-1\right)=-\left(-\frac{100}{99}\right)=\frac{100}{99}\)Câu trả lời: \(a,A=\frac{1}{2}-\frac{1}{3}=\frac{1}{6};B=\frac{1}{3}-\frac{1}{4}=\frac{1}{12};C=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\)\(b,A+B=\frac{1}{6}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4};A+B+C=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{10+5+3}{60}=\frac{18}{60}=\frac{3}{10}\)\(c,D=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)\(E=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)\(=-\left(-\frac{1}{99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+\frac{1}{97\cdot96}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)\(=-\left(-\frac{1}{99}+\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)\(=-\left(-\frac{1}{99}-1\right)=-\left(-\frac{100}{99}\right)=\frac{100}{99}\)

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Phạm Anh Tùng

15 tháng 9 2021 - olm

giup mik đi mà

#Toán lớp 7

Athanasia Karrywang

15 tháng 9 2021

\(a,A=\frac{1}{2}-\frac{1}{3}=\frac{1}{6};B=\frac{1}{3}-\frac{1}{4}=\frac{1}{12};C=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\)

\(b,A+B=\frac{1}{6}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4};A+B+C=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{10+5+3}{60}=\frac{18}{60}=\frac{3}{10}\)

\(c,D=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(E=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(=-\left(-\frac{1}{99}+\frac{1}{99\cdot98}+\frac{1}{98\cdot97}+\frac{1}{97\cdot96}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)

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

\(=-\left(-\frac{1}{99}-1\right)=-\left(-\frac{100}{99}\right)=\frac{100}{99}\)

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