Tính: 1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90-1/3-1/15-1/35-1/63-1/99-1/143
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Những câu hỏi liên quan
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
HC
2
25 tháng 11 2015
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{11.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=1-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{57}{130}\)
25 tháng 11 2015
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}-\frac{1}{3}-\frac{1}{15}-.....-\frac{1}{143}\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{143}\right)\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{11.13}\right)\)
\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)\(=\left(\frac{1}{1}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{117}{130}-\frac{78}{130}=\frac{39}{130}=\frac{3}{10}\)
NT
0
30 tháng 6 2020
\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{5}{39}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
9 tháng 6 2016
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
9 tháng 6 2016
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
mk nha cac ban
25 tháng 8 2016
\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}\)
\(=\frac{-79}{90}\)
25 tháng 8 2016
1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2
= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)
= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)
= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)
= 1/90 - (1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
29 tháng 8 2023
a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
HM
8 tháng 9 2016
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
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-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130