a,1/ 1 x2 + 1/ 2x3 + 1/ 3x4 + ...... + 1/ 98x99 +1/99x100
b, 12 +1/4 + 1/8 + 1/16 +1/64 +1/128
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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=\(\frac{99}{100}\)
XO
3 tháng 7 2019
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 = \(\frac{1}{2}-\frac{1}{2^9}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4
Câu a:
A = \(\frac{1}{2\times3}\) + \(\frac{1}{3\times4}\) + \(\frac{1}{4\times5}\) + \(\frac{1}{5\times6}\) + \(\frac{1}{6\times7}\) + \(\frac{1}{7\times8}\)
A = \(\frac12-\frac13\) + \(\frac13-\frac14\) + \(\frac14-\frac15\) + \(\frac15-\frac16\) + \(\frac16-\frac17\) + \(\frac17-\frac18\)
A = \(\frac12-\frac18\)
A = \(\frac38\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 4
Câu b:
A = \(\frac12\) + \(\frac14\) + \(\frac18\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + \(\frac{1}{64}\) + \(\frac{1}{128}\) + \(\frac{1}{256}\)
2 x A = 1 + \(\frac12\) + \(\frac14\) + \(\frac18\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + \(\frac{1}{64}\) + \(\frac{1}{128}\)
2 x A - A = 1 + \(\frac12\) +\(\frac14\) + \(\frac18\) + \(\frac{1}{16}\) + \(\frac{1}{32}\) + \(\frac{1}{64}\) + \(\frac{1}{128}\) - \(\frac12-\frac14\) -...-\(\frac{1}{128}\) -\(\frac{1}{256}\)
A x (2 - 1) = (1 - \(\frac{1}{256}\)) + (\(\frac12\)-\(\frac12\)) +...+(\(\frac{1}{128}\) - \(\frac{1}{128}\))
A = 1 - \(\frac{1}{256}\) + 0 + 0+...+ 0
A = \(\frac{255}{256}\)
TP
0
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 :))
NM
3
VH
28 tháng 6 2016
1/1.2 +1/2.3 +1/3.4 +...+1/98.99 +1/99.100
=1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100
=1-1/100=100/100-1/100=99/100
28 tháng 6 2016
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow1-\frac{1}{100}=\frac{99}{100}\)
NM
15
23 tháng 2 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
LH
5 tháng 8 2016
Cho hai số biết rằng bớt số thứ nhất 28 đơn vị thì được số thứ hai va 1/3 số thứ nhất bằng 3/5 số thứ hai.Tìm hai số đó
19 tháng 3 2016
ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
=> \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
DN
10
10 tháng 4 2015
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)
13 tháng 8 2016
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)
\(\) a) Đặt \(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots+\frac{1}{98\times99}+\frac{1}{99\times100}\)
\(A=\frac11-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac11=\frac{1}{100}=\frac{99}{100}\)
b) Đặt \(B=\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2\times B=\left(\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\times2\)
\(2B=1+\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2B-B=\left(1+\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(B=1-\frac{1}{128}=\frac{127}{128}\)
A=11−21+21−31+31−41+⋯+981−991+991−1001
\(A = \frac{1}{1} = \frac{1}{100} = \frac{99}{100}\)
b) Đặt \(B = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128}\)
\(2 \times B = \left(\right. \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} \left.\right) \times 2\)
\(2 B = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}\)
\(2 B - B = \left(\right. 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} \left.\right) - \left(\right. \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} \left.\right)\)
\(B = 1 - \frac{1}{128} = \frac{127}{128}\)