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Khuất Đình Trung
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Lê Bảo Ngân
1 tháng 5 2022 lúc 21:05

81

noname
1 tháng 5 2022 lúc 21:42

\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+....+1-\dfrac{1}{90}=1+1+...+1-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)=9-\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{9x10}\right)=9-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)=9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)

Tran Tuyet
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Hquynh
20 tháng 6 2023 lúc 10:38

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}\right)\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8-1}{8}=\dfrac{7}{8}\)

『Kuroba ム Tsuki Ryoo...
20 tháng 6 2023 lúc 10:38

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

`=`\(1-\dfrac{1}{8}\)

`=`\(\dfrac{7}{8}\)

Tran Tuyet
20 tháng 6 2023 lúc 10:38

sos 

 

 

Quang Trí Lâm
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Hatsune Miku
10 tháng 4 2017 lúc 12:30

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)\(+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)\(+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

Trịnh Văn Đại
10 tháng 4 2017 lúc 12:30

1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72

=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9

=1-1/2+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

=1-1/9

=8/9

Nguyễn Tiến Dũng
11 tháng 7 2017 lúc 8:40

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

anh hói
Xem chi tiết

A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{812}\) + \(\dfrac{1}{870}\)

A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{28\times29}\)\(\dfrac{1}{29\times30}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)  - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+\(\dfrac{1}{28}\)-\(\dfrac{1}{29}\)\(\dfrac{1}{29}\) - \(\dfrac{1}{30}\)

A = 1 - \(\dfrac{1}{30}\)

A = \(\dfrac{29}{30}\)

Đỗ Bảo Châu
15 tháng 8 2023 lúc 20:29

29/30 nha

Đoàn Quỳnh Lâm
Xem chi tiết
Ko cần bt
5 tháng 8 2021 lúc 17:26

đề thiếu bn ơi

Khách vãng lai đã xóa
ĐINH ĐÔNG QUÝ
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Vương Hải Nam
11 tháng 4 2019 lúc 19:50

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(S=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(S=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(S=\frac{1}{3}-\frac{1}{8}\)

\(S=\frac{8}{24}-\frac{3}{24}\)

\(S=\frac{5}{24}\)

Huỳnh Quang Sang
11 tháng 4 2019 lúc 19:50

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(S=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(S=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(S=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{3}-\frac{1}{8}\)

\(=\frac{8}{24}-\frac{3}{24}\)

\(=\frac{5}{24}\)

jksfhisd
Xem chi tiết
Xyz OLM
7 tháng 6 2019 lúc 14:43

1)

A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

   = \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)

   = \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)

   = \(\frac{1}{5}-\frac{1}{12}\)

   = \(\frac{7}{60}\)

B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)

   = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

   = \(\frac{3.4.5.....100}{2.3.4....99}\)

   = \(\frac{100}{2}=50\)

C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)

   = \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)

   = \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)

   = \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)

   = \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)

   

Nguyễn Vũ Minh Hiếu
7 tháng 6 2019 lúc 17:45

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

~ Hok tốt ~

Vũ Huỳnh Phong
8 tháng 6 2019 lúc 8:11

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

....

nguyenthimyhong
Xem chi tiết
TFBoys_Châu Anh
15 tháng 4 2016 lúc 18:24

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\)

CHÚC BN HỌC TỐT !!! ^_^

Ngo Lan Huong
Xem chi tiết
Trần Xuân Thành
26 tháng 1 2016 lúc 10:15


 

=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7

=1-1/7

=6/7

nguyễn quang vinh
26 tháng 1 2016 lúc 10:17

\(\frac{113}{140}\frac{ }{ }\)

Võ Thạch Đức Tín 1
26 tháng 1 2016 lúc 10:23

=\(\frac{6}{7}\)