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1/6+1/12+1/20+1/30+1/42+1/56
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Những câu hỏi liên quan
31 tháng 5 2015
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\)
12 tháng 7 2017
S= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
S= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 -1/6 +1/6 - 1/7 + 1/7 - 1/8
S= 1/2 - 1/ 8
S= 3/8
12 tháng 7 2017
S= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
= 1/2 - 1/3 + 1/3 - ...+ 1/7 - 1/8
= 1/2 - 1/8
= 3/8
22 tháng 4 2015
A=1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=1/2-1/8=3/8
22 tháng 4 2015
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)\(+\frac{1}{7}-\frac{1}{8}\)
=\(\left(\frac{1}{2}-\frac{1}{8}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{7}-\frac{1}{7}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{8}\right)+0+...+0\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\)
NV
3
4 tháng 5 2015
A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8
= 1/2 - 1/3 + 1/3 -1/4 + 1/4 - 1/5 + 1/5 - 1/6 +1/6 - 1/7 + 1/7 - 1/8
= 1/2 - 1/8
= 3/8
4 tháng 5 2015
A=1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
=1/2-1/3+1/4-1/5+1/5-1/6+1/7-1/8
=1/2-1/8
=4/8-1/8
=3/8
MT
1
IW
13 tháng 9 2016
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{56}\)
\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{7.8}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A=1-\frac{1}{8}\)
\(\Rightarrow A=\frac{7}{8}\)
PV
1
5 tháng 10 2021
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
JG
2
13 tháng 6 2020
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{5.4}+\frac{6-5}{5.6}+\frac{7-6}{7.6}+\frac{8-7}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{3}{8}\)
BT
30 tháng 3 2018
a/ \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}=\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}=\)
= 1-1/8 = 7/8
b/ = (17/9+1/9)+(19/13+7/13)+(14/6+10/6) = 18/9 + 26/13 + 24/6 = 2+2+4 = 8
Theo đề ra, ta có:
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(=\dfrac{1}{2}+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)-\dfrac{1}{8}\)
\(=\dfrac{1}{2}+0+0+0+0+0-\dfrac{1}{8}\)
\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{4}{8}-\dfrac{1}{8}=\dfrac{3}{8}\)
Chúc bạn học tốt!
Đặt \(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(A=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
\(A=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)