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NH
6
ML
6 tháng 8 2017
\(a,\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+....+\frac{4}{16.18}+\frac{4}{18.20}\)
\(=\frac{4}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(=2.\frac{9}{20}\)
\(=\frac{9}{10}\)
\(b,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
WR
8 tháng 7 2015
2[1/2X4+1/4X6+1/6X8+...+1/Xx(X+2)]=11/45x2
2/2x4+2/4x6+2/6x8+....+2/Xx(X+2)=22/45
1/2-1/4+1/4-1/6+1/6-1/8+...+1/x-1/x+2=22/45
1/2-1/x+2=22/45
1/x+2=1/2-22/45
1/x+2=1/90
=>x+2=90
=>x=88
vậy x=88
8 tháng 7 2015
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{45}\)
\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{x\left(x+2\right)}=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+2}=\frac{22}{45}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2}-\frac{22}{45}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{90}\)
=>x+2=90
=>x=90-2
=>x=88
vậy x=88
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
26 tháng 6 2023
A = \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+\(\dfrac{5}{16.21}\)+...+\(\dfrac{5}{101.106}\)
A = \(\dfrac{1}{1}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{106}\)
A = \(\dfrac{105}{106}\)
B = \(\dfrac{3}{1.4}\) +\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{97.100}\)
B = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
B = \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
C = \(\dfrac{1}{2.7}+\dfrac{1}{7.12}\) + \(\dfrac{1}{12.17}\)+...+ \(\dfrac{1}{97.102}\)
C= \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{5}{2.7}+\dfrac{5}{7.12}+\dfrac{5}{12.17}+...+\dfrac{5}{97.102}\))
C = \(\dfrac{1}{5}\)\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{17}\)+...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{102}\))
C = \(\dfrac{1}{5}\) \(\times\) \(\dfrac{25}{51}\)
C = \(\dfrac{5}{51}\)
D = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)
D = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)
D = \(\dfrac{1}{1}\) - \(\dfrac{1}{9}\)
D = \(\dfrac{8}{9}\)
E = \(\dfrac{3}{2.4}\)+\(\dfrac{3}{4.6}\)+\(\dfrac{3}{6.8}\)+...+\(\dfrac{3}{98.100}\)
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{98.100}\))
E = \(\dfrac{3}{2}\)\(\times\)( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))
E = \(\dfrac{3}{2}\) \(\times\) \(\dfrac{49}{100}\)
E = \(\dfrac{147}{200}\)
CQ
0
KK
21 tháng 2 2017
TA có
4-2/2*4+6-4/4*6+8-6/6*8+...+2016-2014/2014*2016
=1/2-1/4+1/4-1/6+...+1/2014-1/2016
=1/2+1/4-1/4+1/6-1/6+...+1/2014-1/2014-1/2016
=1/2-1/2016
=1007/2016
6 tháng 7 2015
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
19 tháng 3 2017
\(\frac{1}{2x4}\)+ \(\frac{1}{4x6}\)+ ... + \(\frac{1}{98x100}\)= \(\frac{1}{2}\)x(\(\frac{4-2}{2x4}\)+\(\frac{6-4}{4x6}\)+ ... + \(\frac{100-98}{98x100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{8}\)+ ... + \(\frac{1}{98}\)-\(\frac{1}{100}\))
= \(\frac{1}{2}\)x(\(\frac{1}{2}\)-\(\frac{1}{100}\)) = \(\frac{49}{200}\)
16 tháng 3 2016
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(D=\frac{6}{2x4}+\frac{6}{4x6}+\frac{6}{6x8}+....+\frac{6}{48x50}\)
\(=\frac{6}{2}x\left(\frac{2}{2x4}+\frac{2}{4x6}+\frac{2}{6x8}+....+\frac{2}{48x50}\right)\)
\(=3x\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{48}-\frac{1}{50}\right)\)
\(=3x\left(\frac{1}{2}-\frac{1}{50}\right)=3x\frac{12}{25}=\frac{36}{25}\)
Vậy D=36/25
D=6/2x4 + 6/4x6 + 6/6x8 + ...+ 6 /48 x50
D=3 x (2/2x4 + 2/4x6 + 2/6x8 + ...+ 2 /48 x50)
D= 3x (1/2 - 1/4 + 1/4 - 1/6 + 1/6-1/8 + ... + 1/48 - 1/50)
D= 3 x (1/2 - 1/50)
D= 3 x 12/25
D= 36/25