1/2+5/6+11+12+19/20+29/30+41/42+55/56
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VT
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
4 tháng 9 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{5}{6}\) + \(\dfrac{11}{12}\) + \(\dfrac{19}{20}\) + \(\dfrac{29}{30}\) + \(\dfrac{41}{42}\) + \(\dfrac{55}{56}\)
A = (1 - \(\dfrac{1}{2}\)) + ( 1 - \(\dfrac{1}{6}\)) + (1 - \(\dfrac{1}{12}\)) + (1 - \(\dfrac{1}{20}\)) +(1-\(\dfrac{1}{30}\))+(1-\(\dfrac{1}{42}\))+(1-\(\dfrac{1}{56}\))
A = (1 + 1+1 + 1 + 1+1+1)- (\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\))
A = 7 - (\(\dfrac{1}{1\times2}\)+\(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\))
A = 7 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\))
A = 7 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{8}\))
A = 7 - \(\dfrac{7}{8}\)
A = \(\dfrac{49}{8}\)
DD
Đoàn Đức Hà
Giáo viên
2 tháng 6 2021
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
KT
15 tháng 10 2018
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{38}{5}\)
BN
2
16 tháng 5 2022
`= 1 - 1/2 + 1 - 1/6 + ... + 1 - 1/56`
`= 1 - 1/(1.2) + 1 - 1/(2.3) + ... + 1 - 1/(7.8)`
`= 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 + 1/7 - 1/8`.
`= 8 - 1/8`
`= 63/64`.
2
16 tháng 5 2022
`A=1/2+5/6+11/12+19/20+29/30+41/42+55/56`
`A=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56`
`A=(1+1+1+1+1+1+1)-(1/2+1/6+1/12+....+1/56)`
`A=7-(1/[1xx2]+1/[2xx3]+1/[3xx4]+....+1/[7xx8])`
`A=7-(1-1/2+1/2-1/3+1/3-1/4+....+1/7-1/8)`
`A=7-(1-1/8)`
`A=7-(8/8-1/8)`
`A=7-7/8`
`A=56/8-7/8=49/8`
14 tháng 5 2016
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72=(1-1/2)+(1-1/6)+(1-1/12)+(1-1/20)+(1-1/30)+(1-1/42)+(1-1/56)+(1-1/72)=(1+1+1+1+1+1+1+1)-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72)=8-(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)=8-(1-1/2+1/2-1/3+...+1/8-1/9)=8-(1-1/9)=8-8/9=72/9-8/9=64/9
7 tháng 6 2016
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72
=10-(1/2+1/6+..+1/110)
=10-(1/1x2+1/2x3+...+1/10x11)
=10-(1-1/2+1/2-1/3+...+1/10-1/11)
=10-(1-1/11)
=10-10/11
=100/11
18 tháng 6 2021
`A=1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90`
`=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90`
`=9-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`
`=9-(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))`
`=9-(1-1/2+1/2-1/3+.....+1/9-1/10)`
`=9-(1-1/10)`
`=9-1+1/10=8+1/10=81/10`
18 tháng 6 2021
A = \(\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{90}\right)\)
= \(9-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\)
=\(9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
= \(9-\left(1-\dfrac{1}{10}\right)\)
= \(9-\dfrac{9}{10}=\dfrac{81}{10}\)
16 tháng 11 2020
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
Đề bài phải là:
1/2 + 5/6 + 11/12 + 19/20 + 41/42 + 55/56 chứ em
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