NH
1
K
Khách
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 3 2023
A = \(-\dfrac{1}{20}\) + \(\dfrac{-1}{30}\) + \(\dfrac{-1}{42}\) + \(\dfrac{-1}{56}\) + \(\dfrac{-1}{72}\) + \(\dfrac{-1}{90}\)
A = - ( \(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\))
A = - ( \(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\))
A = - ( \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\))
A = - (\(\dfrac{1}{4}-\dfrac{1}{10}\))
A = - \(\dfrac{3}{20}\)
TH
3 tháng 3 2020
\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(D=\frac{1}{90}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\right)\)
\(D=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(D=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(D=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
3 tháng 3 2020
D=1/90 - 1/72 -1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
D=1/90-(1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2)
D=1/90-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72)
D=1/90-(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)
D=1/90-(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)
D=1/90-(1/1-1/9)
D=1/90-8/9
D=(-79/90)
PT
2
N
0
DN
2
A
27 tháng 2 2020
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{4}-\frac{1}{10}=\frac{6}{40}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7 2023
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
25 tháng 11 2015
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-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130
NT
1
TN
7 tháng 5 2016
\(M=\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=8+\frac{1}{10}\)
\(=\frac{81}{10}\)
D=\(-\dfrac{1}{4.5}\)+(\(-\dfrac{1}{5.6}\))+(\(-\dfrac{1}{6.7}\))+(\(-\dfrac{1}{7.8}\))+(\(-\dfrac{1}{8.9}\))+(\(-\dfrac{1}{9.10}\))
D=\(-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
D=\(-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
D=\(-\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)
D=\(-\dfrac{3}{20}\)