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Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Ta có :
\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{13.14}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)
\(C=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(1-\frac{1}{14}\right)\)
\(C=1-\frac{1}{14}\)
\(C=\frac{14}{14}-\frac{1}{14}\)
\(C=\frac{14-1}{14}\)
\(C=\frac{13}{14}\)
Vậy \(C=\frac{13}{14}\)
Chúc bạn học tốt ~
\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{13\cdot14}\)
\(C=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+....+\frac{14-13}{13\cdot14}\)
\(C=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+....+\frac{14}{13\cdot14}-\frac{13}{13\cdot14}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{13}-\frac{1}{14}\)
\(C=1-\frac{1}{14}\)
\(C=\frac{13}{14}\)
dấu "." là dấu nhân nhs
Ta có: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)=\frac{2.19}{20}=\frac{19}{10}\)
\(\frac{2}{1\times2}+\frac{2}{2\times3}+......+\frac{2}{19\times20}\)
\(=2\left(\frac{1}{1\times2}+\frac{1}{2\times3}+.......+\frac{1}{19\times20}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
k cho mình nha bạn
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
1/1x2+1/2x3+1/3x4+..+1/9x10
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...-1/10
=1-1/10
=9/10
1/1x2 +1/2x3 +1/3x4 +1/4x5 +1/5x6
= 1/1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
=1/1 - 1/6
=5/6
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1/2+(1/3-1/3)+(1/4-1/4)+(1/5-1/5)-1/6
=1/2-1/6
=4/12=1/3
chuc ban hoc tot ^-^
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\)\(1-\frac{1}{2014}\)
\(=\)\(\frac{2014}{2014}-\frac{1}{2014}\)
\(=\)\(\frac{2013}{2014}\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{2013}{2014}\)
Dấu \(.\) là dấu nhân nhé
Chúc bạn học tốt ~
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2013\times2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}\)
\(=\frac{2013}{2014}\)
CHÚC BN HỌC TỐT!!!!!