5/6 : 1/6% - 4 % = ... ( SỐ THẬP PHÂN )
4/2X4 + 4/6X8 ....... 4/180X182 =
a) 5/6 : 1/12% - 5 % = ... ( số thập phân )
b) 2/2x4 + 2/6x8 + ... 2/188 x 190 = .... ( phân số )
b)\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{188}-\frac{1}{190}\)
\(=\frac{1}{2}-\frac{1}{190}\)
\(=\frac{47}{95}\)
tính : a, 4/2x4+4/4x6+4/6x8+......+4/16x18+4/18x20
b, 1/2+1/6+1/12+1/20+......+1/90
\(a,\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{16.18}+\frac{4}{18.20}\)
\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=2.\frac{9}{20}\)
\(=\frac{9}{10}\)
\(b,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
a, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+..+\frac{4}{16.18}+\frac{4}{18.20}\)
\(=\frac{4}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=2\cdot\frac{9}{20}=\frac{9}{10}\)
b, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
( 1/2x4+1/4 x 6+1/6x8+....+1/98 x 100) / y = 147/50 tìm y
\(\frac{\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}}{y}=\frac{147}{50}\)
\(\frac{\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{98\times100}\right)}{y}=\frac{147}{50}\)
\(\frac{\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)}{y}=\frac{147}{50}\)
\(\frac{\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{100}\right)}{y}=\frac{147}{50}\)
\(\frac{\frac{1}{2}\times\frac{49}{100}}{y}=\frac{147}{50}\)
\(\frac{49}{200}:y=\frac{147}{50}\)
\(y=\frac{49}{200}:\frac{147}{50}\)
\(y=\frac{1}{12}\)
4/ 2x4 +4/4x6+4/6x8+.......+4/2008x2010
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=\frac{2.2004}{2010}=\frac{2004}{1005}\)
\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)
\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1004\cdot1005}\right)\)
\(=2\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
\(=2\cdot\left(1-\frac{1}{1005}\right)=2\cdot\frac{1004}{1005}=\frac{2008}{1005}\)
4/2x4 + 4/4x6 + 4/6x8 + .... + 4/2014x 2016
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2014.2016}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)=2.\left(\frac{1008}{2016}-\frac{1}{2016}\right)=2.\frac{1007}{2016}=\frac{1007}{1008}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{2014.2016}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2014.2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2014}-\frac{1}{2016}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=2.\frac{1007}{2016}\)
\(=\frac{1007}{1008}\)
4/2x4+4/4x6+4/6x8+...+4/18 x20 = ?
Gọi tổng là A ta có :
A x 2 = 2/2.4 + 2/4.6 + 2/6.8 + ... + 2/18.20
A x 2 = 1/2 - 1/4 - 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/18 - 1/20
A x 2 = 1/2 - 1/20
A x 2 = 9/20
A = 9/20 : 2
A = 9/40
tính nhanh 4/2x4 + 4/4x6 + 4/6x8 + .... + 4/2008x 2010
Đặt:A = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
=> A = 2.(\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+.....+\frac{2}{2008.2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{2008}-\frac{1}{2010}\)
=> A = 2.(\(\frac{1}{2}-\frac{1}{2010}\))
=> A = 2.\(\frac{502}{1005}\)
=> A = \(\frac{1004}{1005}\)
đặt A= \(\frac{4}{2.4}+\frac{4}{4.6}+...+\frac{4}{2008.2010}\)
=> 1/2.A=\(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\)
= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\)
=\(\frac{1}{2}-\frac{1}{2010}\)
=\(\frac{502}{1005}\)
Vậy biểu thức cần tìm có giá trị là \(\frac{502}{1005}\)
nhầm nha
=> A= \(\frac{502}{1005}:\frac{1}{2}\)=\(\frac{1004}{1005}\)
K=4/2x4+4/4x6+4/6x8+...+4/2008x2010
F=1/18+1/54+1/108+...+1/990
K = đề bài
= 2 . ( 2/2.4 + 2/4.6 + 2/6.8 + . . . + 2/2008.2010 )
= 2 . ( 1 - 1/4 + 1/4 - 1/6 + 1/8 - 1/8 + . . . + 2/2008 - 2/2010 )
= 2 . ( 1 - 2/2010 )
= ( phần còn lại bạn tự tính nha )
k cho mình đó, bài này mình làm rồi nên đúng 100% lun, sorry nha mình ngại viết nhiều
a) \(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(\Leftrightarrow K=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(\Leftrightarrow K=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Leftrightarrow K=2.\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow K=2.\frac{2009}{2010}=\frac{2009}{1005}\)
b) F=1/18 + 1/54 + 1/108 +...+ 1/990
=> \(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(\Leftrightarrow F=3.\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\right)\)
\(\Leftrightarrow F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(\Leftrightarrow F=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(\Leftrightarrow F=1-\frac{1}{33}=\frac{32}{33}\)
Cho B = 4\2x4 + 4\4x6 + 4\6x8 + ..... 4\98x100
Tính 5B ?
\(B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{98.100}\)
\(\Rightarrow5B=\dfrac{20}{2.4}+\dfrac{20}{4.6}+...+\dfrac{20}{98.100}\)
\(\Rightarrow5B=10\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)
\(\Rightarrow5B=10.\dfrac{49}{100}\)
\(\Rightarrow5B=\dfrac{49}{10}\)
Vậy \(5B=\dfrac{49}{10}\)
Ta có: B = \(\dfrac{4}{2.4}\) + \(\dfrac{4}{4.6}\) + \(\dfrac{4}{6.8}\) + ... + \(\dfrac{4}{98.100}\).
=> \(\dfrac{B}{2}\) = \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) + \(\dfrac{2}{6.8}\) + ... + \(\dfrac{2}{98.100}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\) + ... + \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\)
= \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\) = \(\dfrac{49}{100}\).
=> B = \(\dfrac{49}{200}\).
=> 5B = \(\dfrac{49}{200}\) . 5 = \(\dfrac{49}{40}\).
Vậy 5B = \(\dfrac{49}{40}\).