hi bn

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)

=1-1/2+1/2-1/3+......+1/5-1/6

=1-1/6

=5/6

Tick

A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2018.2019}\)

A= 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2018}-\frac{1}{2019}\)

A= 1 - \(\frac{1}{2019}\)

A= \(\frac{2018}{2019}\)

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2018\cdot2019}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(A=1-\frac{1}{2019}\)

\(=\frac{2018}{2019}\)

Vậy \(A=\frac{2018}{2019}\)

HOK TỐT ==.==

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{2018.2019}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{2019}\)

\(\Rightarrow A=1-\frac{1}{2019}\)

\(\Rightarrow A=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)

Vậy A = \(\frac{2018}{2019}\)

S=1-1/100=99/100

bạn tách ra, 1/1.2=1-1/2 cứ như thế, rồi trừ đi còn 1-1/100=99/100

S = 99/100

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}\)

\(A=\frac{5}{6}\)

_Chúc bạn học tốt_

Ta có : \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(=1-\frac{1}{5}=\frac{4}{5}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}_{ }-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}\)

\(A=\frac{5}{6}\)

49/50 day ban a k cho minh nha

bạn ơi đánh có dấu đi mình ko hiểu ý bạn là gì

Tại sao ra kết quả như vậy hả bạn ???

=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=1-\frac{1}{6}=\frac{5}{6}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2009\cdot2010}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(=\frac{1}{1}-\frac{1}{2010}\)

\(=\frac{2010}{2010}-\frac{1}{2010}\)

\(=\frac{2009}{2010}\)