bài 1 : 1/1x2+1/2x3+1/3x4+1/5x6+1/6x7
Bài 2:1/2+1/6+1/12+1/30+1/42+1/56+1/72
Bài 3: (1-1/2)x(1-1/3)x....x(1-2019)
Bài 4:1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
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8 tháng 8 2019
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)
\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)
\(< =>\frac{255}{256}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(< =>\frac{1}{1}-\frac{1}{100}\)
\(< =>\frac{99}{100}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(< =>\frac{1}{100}\)
mk chuc ban hoc tot nhe :))
ND
0
DQ
2
LM
19 tháng 8 2016
A=1-1/2+1/2-1/3+1/3-1/4+...+1/64-1/128
A=1-1/128
A=127/128
Vậy A=\(\frac{127}{128}\)
LM
19 tháng 8 2016
B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
B=1-1/100
B=99/100
Vậy B=\(\frac{99}{100}\)
8 tháng 8 2017
a) = \(\frac{127}{96}\)
b) = \(\frac{255}{256}\)
c) Mik bỏ nha
d) = \(\frac{1023}{512}\)
e) = \(\frac{2343}{625}\)
DA
16 tháng 8 2017
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
1) 1/1.2 + 1/2.3 + ... + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7
= 1 - 1/7
= 6/7
2) 1/2 + 1/6 + 1/12 + .. + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9
= 1 - 1/9
= 8/9
3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)
= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)
= \(\frac{1.2....2019}{2.3...2020}\)
= \(\frac{1}{2020}\)
4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)
= \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)
=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)
A = \(\frac{1}{2}-\frac{1}{2^9}\)