\(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(B=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
Tính giá trị của biểu thức:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A=1-\frac{1}{2010}\)
\(A=\frac{2009}{2010}\)
K=\(\frac{4}{2.4}\)+\(\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
F=\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
I=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(K=2\times\frac{502}{1005}\)
\(K=\frac{1004}{1005}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{10}{33}:3\)
\(F=\frac{10}{99}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}\)
\(I=\frac{2009}{2010}\)
Tính:
a, I = \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ...+ \(\frac{1}{2009.2010}\)
b, K = \(\frac{4}{2.4}\)+ \(\frac{4}{4.6}\)+ \(\frac{4}{6.8}\)+...+ \(\frac{4}{2008.2010}\)
c, F = \(\frac{1}{18}\)+ \(\frac{1}{54}\)+ \(\frac{1}{108}\)+...+ \(\frac{1}{990}\)
a) \(I=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2009\cdot2010}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)
b) \(K=\frac{4}{2\cdot4}+\frac{4}{2\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\)
\(\frac{1}{2}K=\frac{1}{2}\left(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+....+\frac{4}{2008\cdot2010}\right)\)
\(\frac{1}{2}K=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\)
\(\frac{1}{2}K=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2008}-\frac{2}{2010}\)
\(\frac{1}{2}K=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(K=\frac{2009}{2010}:\frac{1}{2}=\frac{2009}{1005}\)
Tính :
a)\(\left[-\frac{1}{2}\right]^2:\frac{1}{4}-2\left[-\frac{1}{2}\right]^2\)
b)\(-\frac{7}{37}.1\frac{3}{8}-4\frac{5}{8}.\frac{7}{12}+\frac{1}{2}\)
c)\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
d)\(H=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
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(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\))
A=2(\(\frac{1}{2}-\frac{1}{2010}\))
A=2.\(\frac{502}{1005}\)
A=\(\frac{1004}{1005}\)
Mình ko ghi lai đề nha
4/2.4/4+4/4.4/6+......+4/2008.4/2010=4/2.4/2010=4/1005
Mình ko bt đúng ko nữa nha
A x 2/4= 2/2x4 + 2/4x6 + 2/6x8 +............+ 2/2008x2010
A x 2/4=4-2/2x4 + 6-4/4x6 + 8-6/6x8 +.......+ 2010-2008/2008x2010
A x 2/4=4/2x4 - 2/2x4 + 6/4x6 - 4/4x6 +8/6x8 -6/6x8 +............+ 2010/2008x2010 - 2008/2008x2010
A x 2/4=1/2-1/2010
A x 2/4=502/1005
A= 502/1005 / 2/4
A=1004/1005
a\(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\)
b \(F=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
a,\(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\) =\(\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\) =\(\left(3-\frac{31}{12}\right):\left(4-\frac{1}{3}\right)\) =\(\frac{5}{12}:\frac{11}{3}\) =\(\frac{5}{44}\) b, F=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.......+\frac{4}{2008.2010}\) =\(2.\left(1-\frac{2}{2010}\right)\) =\(2.\frac{1004}{1005}\) =\(\frac{2008}{1005}\)
a) \(\left(3-2\frac{1}{3}+\frac{1}{4}\right):\left(4-5\frac{1}{6}+2\frac{1}{4}\right)\)
\(=\left(3-\frac{7}{3}+\frac{1}{4}\right):\left(4-\frac{31}{6}+\frac{9}{4}\right)\)
\(=\frac{3-\frac{7}{2}+\frac{1}{3}}{\frac{13}{12}}\)
\(=\frac{11}{13}\)
\(F=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(F=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(F=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(F=2.\frac{502}{1005}\)
\(F=\frac{1004}{1005}\)
C=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{2008.2010}\)
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2010}\right)\) \(;C=\frac{1}{2}.\frac{502}{1005}=\frac{251}{1005}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{1004.1005}\)
=\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1004.1005}\right)\)
=\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\right)\)
=\(2\left(1-\frac{1}{1005}\right)\)
=\(2.\frac{1004}{1005}\)
=\(\frac{2008}{1005}\)
P/s: Không biết đúng không nữa, làm đại ^.^
Ta thấy : \(\frac{4}{2.4}=\frac{1}{2}\left(\frac{4}{2}-\frac{4}{4}\right);\frac{4}{4.6}=\frac{1}{2}\left(\frac{4}{4}-\frac{4}{6}\right);...;\frac{4}{2008.2010}=\frac{1}{2}\left(\frac{4}{2008}-\frac{4}{2010}\right)\)
=> C =\(\frac{1}{2}.\left(\frac{4}{2}-\frac{4}{4}+\frac{4}{4}-\frac{4}{6}+\frac{4}{6}-\frac{4}{8}+...+\frac{4}{2008}-\frac{4}{2010}\right)\)
=> C = \(\frac{1}{2}\left(\frac{4}{2}-\frac{4}{2010}\right)=\frac{1}{2}\left(2-\frac{2}{1005}\right)=\frac{1}{2}\left(\frac{2010}{1005}-\frac{2}{1005}\right)\)
=> C = \(\frac{1}{2}\left(\frac{2010-2}{1005}\right)=\frac{1}{2}.\frac{2008}{1005}=\frac{1004}{1005}\)
Tính nhanh
1)A=\(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
2)B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
3)C=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
C : có ở bên dưới rồi, còn A và B thôi
1) A=7(\(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-......+\frac{1}{69}-\frac{1}{70}\) )
A=7 ( \(\frac{1}{10}-\frac{1}{70}\))
A=7x 6/70
A=3/5
Tính nhanh \(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
K = 4/2 - 4/4 + 4/4 - 4/6 + ....... + 4/2008 - 4/2010
K = 4/2 - 4/2010
K = 4016/2010 = 1/1003/1005
\(K=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2002}-\frac{1}{2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(K=2.\frac{502}{1005}\)
\(K=\frac{1004}{1005}\)