tính nhanh:
1/2+1/6+1/12+...+1/90
K
Khách
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Những câu hỏi liên quan
17 tháng 1 2023
sau đây là phần chữa của mình:
\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{3}{10}\)
17 tháng 1 2023
= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{2}{5}\)
NH
2
5 tháng 4 2017
= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ...... + 1/9x10
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4-1/5 +.......+ 1/9 -1/10
= 1/1 - 1/10
= 9/10
VD
14
31 tháng 1 2016
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..........+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{11.12}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{11}-\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
15 tháng 7 2015
A= \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
= \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
15 tháng 7 2015
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
CT
2
13 tháng 9 2023
A=1/20+1/30+1/42+1/56+1/72+1/90
A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=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=1/4-1/10
A=3/20
LN
2
5 tháng 7 2023
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}+\dfrac{1}{110}\)
\(=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}+\dfrac{1}{10x11}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{2}-\dfrac{1}{11}=\dfrac{11}{22}-\dfrac{2}{22}=\dfrac{9}{22}\)
6 tháng 7 2023
1/6+1/12+1/20+1/90+1/110
=1/2x3+1/3x4+1/4x5+...+1/9x10+1/10x11
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-...+1/9-1/10+1/10-1/11
=1/2-1/11=9/22
5 tháng 7 2020
Bài làm
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\frac{1}{1}-\frac{1}{11}\)
\(A=\frac{10}{11}\)
9 tháng 6 2016
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
9 tháng 6 2016
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
mk nha cac ban
HV
2
NA
Nguyễn An Ninh
VIP
13 tháng 5 2023
\(\text{1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90}\)
\(\text{= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.1}\)
\(\text{= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10}\)
\(=1/1-1/10-10/10-1/10-9/10\)
Vậy \(\text{ 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = 9/10}\)
KV
13 tháng 5 2023
Sửa đề:
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{3}{5}\)