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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
6 tháng 9 2016
2;6;12;20;30;42;56;72;90;110
moi so tu 2 den 6 la 4 don vi moi so tiep theo la cong them 2 don vi nhu 2+4=6;6+6=12....
nhu vay 5 so tiep theo cong them vao 110 la ra : so dau tien la: 22 ;24;26;28;30
vay 5 so do la 110 +22=132 ;132 +24 =156 ;156+26 =182;182+28=210 ;210+30=240
tong cac so :2+6+12+20+30+42+56+72+90+110+132+156+182+210+240=1328
ket qua la 1328
chuc ban hoc gioi nha
neu ban cam thay mh dung thi k cho mk nha cac ban !!!!!!! :))) ^^^
DD
Đoàn Đức Hà
Giáo viên
2 tháng 6 2021
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
22 tháng 3 2018
ta có: 1/2=1/1x2
1/6=1/2x3
1/12=1/3x4
1/20=1/4x5
1/30=1/5x6
1/42=1/6x7
1/56=1/7x8
1/72=1/8x9
1/90=1/9x10
1/110=1/10x11
tiếp theo bn tiếp tục nhé
22 tháng 3 2018
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110
=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x1
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11
=1-1/11
=10/11
LH
1
14 tháng 7 2015
A - B = \(\left(1+\frac{1}{2}+1+\frac{1}{12}+1+\frac{1}{30}+1+\frac{1}{56}+1+\frac{1}{90}\right)-\left(1-\frac{1}{6}+1-\frac{1}{20}+1-\frac{1}{42}+1-\frac{1}{72}+1-\frac{1}{110}\right)\)= \(\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)\
= \(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}-5+\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)
NM
3
9 tháng 9 2020
a)A=1/20+1/30+1/42+1/56+1/72+1/90+1/110
= 1/4*5 + 1/5*6 + 1/6*7 + ... + 1/10*11
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/10 - 1/11
= 1/4 - 1/11
= 7/44
b)B=1/2+1/4+1/6+1/8+...+1/512+1/1024
B = 1/2^1 + 1/2^2 + 1/2^3 + ... + 1/2^9 + 1/2^10
2B = 1 + 1/2 + 1/2^2 + ... + 1/2^10 + 1/2^11
2B - B = B = 1 + 1/2^11
\(S=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{110}\)
\(S=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+...+\frac{2}{10\cdot11}\)
\(S=2\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\right)\)
\(S=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(S=2\left(\frac{1}{4}-\frac{1}{11}\right)=2\cdot\frac{7}{44}=\frac{7}{22}\)
thank