A, m= 4/1.3+4/3.5+4/5.7+....+4/2013.2015
b,S=2/5.9+2/9.13+2/13.17+....+2/97.101
c,A=1/5.6+1/6.7+1/7.8+......+1/24.25
d,B=2/1.3+2/3.5+2/5.7+....+2/99.101
e,C=52/1.6+52/6.11+52/11.16+52/16.21+52/21.26+52/26.31
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Những câu hỏi liên quan
10 tháng 4 2016
a, \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
Gọi biểu thức trên là A
17 tháng 2 2020
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
D
2
XO
20 tháng 2 2020
Ta có B = \(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2014}}\)
=> 4B = \(1+\frac{1}{4}+...+\frac{1}{4^{2013}}\)
Lấy 4B trừ B theo vế ta có :
4B - B = \(\left(1+\frac{1}{4}+...+\frac{1}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2014}}\right)\)
=> 3B = \(1-\frac{1}{4^{2014}}\)
=> B = \(\left(1-\frac{1}{4^{2014}}\right):3=\frac{1}{3}-\frac{1}{3.4^{2014}}\)
Lại có C = \(\frac{1}{52}\left(\frac{35}{1.3}+\frac{35}{3.5}+...+\frac{35}{103.105}\right)=\frac{1}{52}.\frac{35}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{103.105}\right)\)
\(=\frac{35}{104}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{103}-\frac{1}{105}\right)\)
\(=\frac{35}{104}.\left(1-\frac{1}{105}\right)=\frac{35}{104}.\frac{104}{105}=\frac{1}{3}\)
Vì \(\frac{1}{3}-\frac{1}{3.4^{104}}< \frac{1}{3}\Rightarrow B< C\)
Vậy B < C
29 tháng 7 2015
a)1/5.6+1/6.7+1/7.8+.......+1/99.100
= (1/5-1/6)+(1/6-1/7)+(1/7-1/8)+.....+(1/99-1/100)
= 1/5 - 1/100
= 19/100
b)2/1.3+2/3.5+2/5.7+.........+2/2013.2015
= (1/1-1/3)+(1/3-1/5)+(1/5-1/7)+.....+(1/2013+1/2015)
= 1/1 - 1/2015
= 2014/2015
MT
29 tháng 7 2015
\(a,\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{5}-\frac{1}{100}=\frac{20}{100}-\frac{1}{100}=\frac{19}{100}\)
\(b,\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\)
\(=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)
3 tháng 4 2018
1/5.6 + 1/6.7 + 1/7.8 +...+ 1/24.25
=1/5 - 1/6 + 1/6-1/7 +1/7-1/8 + ... + 1/24-1/25
=> Kết quả là: 1/5 - 1/25 = 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+...+ 2/99.101
=2/1-2/3 + 2/3-2/5 + 2/5-2/7 + 2/7-2/9 + ... + 2/99-2/101
=> kết quả là 2/1 - 2/101 =200/101
3 tháng 4 2018
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
=\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
=\(\frac{1}{5}-\frac{1}{25}\)
=\(\frac{4}{25}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
=\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
=\(2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
=\(2.\frac{100}{101}\)
=\(\frac{200}{101}\)
7 tháng 6 2016
a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101
= 1 - 1/101
= 100/101
c) 3/1.4 + 3/4.7 + ... + 3/2002.2005
= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005
= 1 - 1/2005
= 2004/2005
d) 5/2.7 + 5/7.12 + ... + 5/1997.2002
= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002
= 1/2 - 1/2002
= 500/1001
7 tháng 6 2016
a,A = \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)
A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)
B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
B=\(1-\frac{1}{101}=\frac{100}{101}\)
c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)
C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)
d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)
D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)
D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)
KT
1
TN
14 tháng 4 2016
bài 1
\(2A=\left(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{99\cdot101}\right)\cdot2\)
\(=5\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\right)\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=5\left(1-\frac{1}{101}\right)\)
\(=5\cdot\frac{100}{101}\)
\(=\frac{500}{101}\Rightarrow A=\frac{500}{101}:2=\frac{250}{101}\)
bài 2:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=-1\)
7 tháng 12 2021
\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{101\cdot106}\right)\\ =5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\\ =5\left(1-\dfrac{1}{106}\right)=5\cdot\dfrac{105}{106}=\dfrac{525}{106}\)
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