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8 tháng 7 2018
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}=\frac{3}{7}\)
8 tháng 7 2018
Đặt \(C=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{84}\)
\(\Rightarrow\frac{C}{2}=1+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{7}\)
\(\Rightarrow C=\left(1+\frac{1}{2}-\frac{1}{7}\right).2\)
TT
3
F
11 tháng 7 2018
Tính nhanh :
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=2\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\frac{1}{10\cdot12}+\frac{1}{12\cdot14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(A=2\cdot\frac{3}{7}\)
\(A=\frac{6}{7}\)
11 tháng 7 2018
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)
\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
_Chúc bạn học tốt_
EL
1
17 tháng 8 2016
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)
8 tháng 6 2023
A = 1/4 x 8 + 1/8 x 12 + 1/12 x 16 + ... + 1/176 x 180
=> 4A = 4/4 x 8 + 4/8 x 12 + 4/12 x 16 + ... + 4/176 x 180
=> 4A = 1/4 - 1/8 + 1/8 - 1/12 + 1/12 - 1/16 + ... 1/176 - 1/180
=> 4A = 1/4 - 1/180
=> 4A = 45/180 - 1/180
=> 4A = 44/180
=> 4A = 11/45
=> A = 11/45 : 4
=> A = 11/180
Vậy A = 11/180
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 6 2023
A = \(\dfrac{1}{4\times8}\) + \(\dfrac{1}{8\times12}\) + \(\dfrac{1}{12\times16}\) +...+ \(\dfrac{1}{176\times180}\)
A = \(\dfrac{1}{4}\) \(\times\)( \(\dfrac{4}{4\times8}\)+ \(\dfrac{4}{12\times16}\)+...+ \(\dfrac{4}{176\times180}\))
A = \(\dfrac{1}{4}\) \(\times\)( \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{16}\) +...+ \(\dfrac{1}{176}\) - \(\dfrac{1}{180}\))
A = \(\dfrac{1}{4}\) \(\times\)(\(\dfrac{1}{4}\) - \(\dfrac{1}{180}\))
A = \(\dfrac{1}{4}\) \(\times\)\(\dfrac{11}{45}\)
A = \(\dfrac{11}{180}\)
KK
4
27 tháng 2 2017
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{96}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+.....+\frac{1}{192}\)
\(\Rightarrow A-\frac{1}{2}A=\frac{1}{3}-\frac{1}{192}\)
\(\Rightarrow\frac{1}{2}A=\frac{21}{64}\)
\(\Rightarrow A=\frac{21}{64}.2=\frac{21}{32}\)
T
3
12 tháng 6 2016
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{96}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{48}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{24}\)
...
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{3}\Rightarrow A=\frac{2}{3}-\frac{1}{96}=\frac{2\cdot32-1}{96}=\frac{63}{96}=\frac{21}{32}\).
A
9 tháng 2 2018
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
9 tháng 2 2018
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)