(1/2.3+1/3.4+1/4.5+...+1/9.10).x2 = 8/5 sos mình cần gấp
Tính hiệu giữa A và B.
A=1/2.3+2/3.4+3/4.5+...+7/8.9+8/9.10
B=1/2+1/3+1/4+1/5+1/6+...+1/9
1/2.3+1/3.4+1/4.5+...+1/1999.2000
giúp mình với tớ cần gấp lắm
\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)
Dạng tổng quát :
\(\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{n\left(n+1\right)}=\frac{1}{2}-\frac{1}{n+1}=\frac{n+1-2}{2\left(n+1\right)}=\frac{n-1}{2\left(n+1\right)}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}\)
\(=\frac{499}{2000}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(=\frac{1}{2}-\frac{1}{2000}\)
\(=\frac{999}{2000}\)
Study well ! >_<
`1/(2*3)+1/(3*4)+1/(4*5)+...+1/(9*10)`
`=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10`
`=1/2-1/10`
`=5/10-1/10`
`=4/10=2/5`
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}\)
\(=\frac{4}{10}\)
\(=\frac{2}{5}\)
ai k tôi k lại
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10
= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
= 1/2 - 1/10
= 5/10 - 1/10
= 4/10 = 2/5
A= 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
A= 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
=> A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
=> A = 1 - 1/10 = 9/10
Vậy A = 9/10
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
A = 1 - 1/10 = 9/10
A= 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
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10
= 1 - 1/10 = 9/10
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
Ta có
\(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{9}-\frac{1}{10}\)
\(=2-\frac{1}{10}\)
\(=\frac{19}{10}\)
Vậy \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)\(=\frac{19}{10}\)
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
\(\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{10}\)
\(=\frac{1}{10}\)
(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-......+1/9-1/10)
1-1/10=9/10
nhớ cho mk
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{2}{5}\)