1+1/1.2+1/2.3+1/3.4+1/4.5+...+1/2006.2007+1/2007.2008
giúp mình với
Cho A=1/1.2+1/2.3+1/3.4+1/4.5+...+1/2018.2019+1/2019.2020 thì A có giá trị là ?
Giúp mình với ạ mình đang cần gấp í:)
\(\text{#}HaimeeOkk\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}\)
\(A=1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{2019}-\dfrac{1}{2019}\right)-\dfrac{1}{2020}\)
\(A=1-0-0-0-...-0-\dfrac{1}{2020}\)
\(A=1-\dfrac{1}{2020}\)
\(A=\dfrac{2019}{2020}\)
Vậy \(A=\dfrac{2019}{2020}\)
Tính tổng giùm mình nhé :1/1.2 +1/2.3 +1/3.4 +1/4.5 +1/5.6 +1/6.7
Ta có : \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)
= 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)
= 1 - \(\frac{1}{7}\)= \(\frac{6}{7}\)
=1-1/2+1/2-1/3+1/3-1/4+...+1/6-1/7=1-1/7=6/7
1/1 -1/2 +1/2 -1/3 +1/3 -1/4 +1/4 -1/5 +1/5 -1/6 +1/6 -1/7
1/1- 1/7 =6/7
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
Làm nhanh giùm mình với ạ,càng chi tiết càng tốt ạ
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{10-1}{10}=\dfrac{9}{10}\)
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=2-1/1.2+3-2/2.3+4-3/3.4+...+10-9/9.10
=1-1/2+1/2-1/3+1/3-1/4+....+1/9-1/10
=1-1/10
=9/10
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
1/1.2+1/2.3+1/3.4+1/4.5+...+1/99.100
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
1/1.2+1/2.3+1/3.4+1/4.5+1/5.6 = ?
=1-1/2+1/2-1/3+......+1/5-1/6
=1-1/6
=5/6
Tick
giúp mìn với 1/1.2 +1/2.3+1/3.4+1/4.5+....+1/[x-1].x+1/x.[x+1]
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
A=1/1.2+1/2.3+1/3.4+1/4.5+....+1/2018.2019
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=1/1.2+1/2.3+1/3.4+1/4.5+....+1/99.100
bạn tách ra, 1/1.2=1-1/2 cứ như thế, rồi trừ đi còn 1-1/100=99/100