S = 2/20 + 2/30 + 2/42 + ... + 2/110
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XO
14 tháng 10 2019
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{10\times11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
VH
14 tháng 10 2019
Đặt A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
=\(\frac{10}{11}\)
Vậy...
HT
4
17 tháng 7 2015
B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + ... + 1/110
B = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/10.110
B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/10 - 1/11
B = 1 - 1/11
B = 10/11
DH
17 tháng 7 2015
B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + .... + 1/110
=1/1x2 + 1/2x3 + 1/ 3x4 + 1/4x5 + 1/5 x6 + 1/ 6x 7 +.....+ 1/10 x11
=1-1/2 + 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7 +......+ 1/10 - 1/11
= 1 + (1/2-1/2)+(1/3-1/3)+(1/4-1/4)+(1/5-1/5)+(1/6-1/6)+.....+(1/10-1/10)-1/11
= 1- 1/11=10/11
Vậy B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + .... + 1/110 = 10/11
MP
1
18 tháng 7 2015
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{10.11}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{8}+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{709}{792}\)
LX
3
20 tháng 8 2017
\(\dfrac{5}{2}+\dfrac{5}{6}+\dfrac{5}{12}+\dfrac{5}{20}+\dfrac{5}{30}+\dfrac{5}{42}+...+\dfrac{5}{110}\)
\(=\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+\dfrac{5}{4.5}+\dfrac{5}{5.6}+\dfrac{5}{6.7}+...+\dfrac{5}{10.11}\)
\(=\dfrac{5}{1}-\dfrac{5}{2}+\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{5}{3}-\dfrac{5}{4}+\dfrac{5}{4}-\dfrac{5}{5}+...+\dfrac{5}{10}-\dfrac{5}{11}\)
\(=\dfrac{5}{1}-\dfrac{5}{11}=\dfrac{50}{11}\)
DH
20 tháng 8 2017
=5(1/2+1/6+1/12+1/20+1/30+1/42+...+1/110)
=5(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/10.11)
=5(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+...+1/10-1/11)
=5(1-1/11)=5.10/11=50/11
NH
15
29 tháng 8 2014
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110= 1 - 1/2 + 1 - 1/6 + 1 - 1/12 .....+1 - 1/110= 10 - ( 1/2 + 1/6 + ...+ 1/110) = 10 - ( 1 - 1/ 2+ 1/2 - 1/ 3+ 1/3 - 1/4 ....+ 1/10 - 1/11)= 10 - (1 - 1/11)= 10 - 10/11 = 100/11
15 tháng 3 2019
\(=\frac{9}{10.11}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10-9}{9.10}-\frac{9-8}{8.9}-...-\frac{2-1}{1.2}\)
\(=\frac{9}{10.11}-\frac{10}{9.10}+\frac{9}{9.10}-...-\frac{2}{1.2}+\frac{1}{1.2}\)
\(=\frac{9}{10.11}-\frac{1}{9}+\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-\frac{1}{7}+\frac{1}{8}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{9}{10.11}+\frac{1}{10}-1\)
\(=-\frac{9}{11}\)
TT
1
B
2
4 tháng 2 2016
a.A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b. 1/2 + 1/6 + 1/12 + … + 1/110
= 1/1.2 + 1/2.3 + 1/3.4 + … + 1/10.11. (dấu . thay dấu x).
= 1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/10 – 1/11
= 1/1 – 1/11
= 10/11
Chúc bạn học giỏi nha!
4 tháng 2 2016
a ) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
Nhân 2 vào hai vế của biểu thức A , ta được :
\(2A=2.\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
Lấy biểu thức 2A - A , ta được :
\(2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^8}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow A=1-\frac{1}{2^9}\Rightarrow A=\frac{512}{512}-\frac{1}{512}=\frac{511}{512}\)
Vậy \(A=\frac{511}{512}\)
b ) Đặt \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(B=1-\frac{1}{11}=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
Vậy \(B=\frac{10}{11}\)
\(S=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{110}\)
\(S=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+...+\frac{2}{10\cdot11}\)
\(S=2\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\right)\)
\(S=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(S=2\left(\frac{1}{4}-\frac{1}{11}\right)=2\cdot\frac{7}{44}=\frac{7}{22}\)
thank