(1.2)2 / ( 2.3)2 . (3.4)2/ (4.5)2 .... (999.1000)2 / (1000.1001)2
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29 tháng 6 2016
\(A=\frac{1^2.2^2.3^2...1000^2}{1.2^2.3^2.4^2...1000^2.1001}=\frac{1}{1001}\)
PT
2
DA
17 tháng 6 2020
12 /1.2 . 22/2.3 . 32/3.4 ... 9992/999.1000
= 1.1/1.2 . 2.2/2.3 . 3.3/3.4........... 999.999/999.1000
= 1/2. 2/3 . 3.4.....999/1000
= 1/1000
L
29 tháng 6 2016
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{100^2}{1000.1001}\)
\(A=\frac{1.1.2.2.3.3.....1000.1000}{1.2.2.3.3.4.....1000.1001}\)
\(A=\frac{\left(1.2.3.....1000\right).\left(1.2.3.....1000\right)}{\left(1.2.3.4....1000\right).\left(2.3.4.....1001\right)}\)
\(A=\frac{1}{1001}\)
DT
29 tháng 6 2016
Ta có A=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{1000^2}{1000.1001}\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{1000}{1001}\)
=\(\frac{1.2.3.....1000}{2.3.4.....1001}\)
=\(\frac{1}{1001}\)
21 tháng 7 2016
S = 2/1×2 + 2/2×3 + 2/3×4 + 2/4×5 + ... + 2/101×102
B = 2 × (1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/101×102)
B = 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/101 - 1/102)
B = 2 × (1 - 1/102)
B = 2 × 101/102
B = 101/51
DD
6
21 tháng 5 2015
\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+........+\frac{2}{8\cdot9}+\frac{2}{9\cdot10}\)
\(=2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+........+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(=2\cdot\frac{9}{10}=\frac{9}{5}\)
đúng nha !
21 tháng 5 2015
Ta có:
2/1.2+2/2.3+2/3.4+2/4.5+2/5.6+2/6.7+2/7.8+2/8.9+2/9.10
=2.(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)
=2.(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)
=2.(1-1/10)
=2.9/10
=9/5
VD
1
Q
27 tháng 10 2016
Ta có: A = 1.2.3+3.4.5+5.6.7+...+99.100.101
A = 1.3 (5-3) + 3.5 (7-3) + 5.7 (9-3) + ............ + 99.101 (103 - 3)
A = (1.3.5 + 3.5.7 + 5.7.9 + .......... + 99.101.103) - (1.3.3 + 3.5.3 + ....... + 99.101.3)
A = (15+99.101.103.105) : 8 - 3.(1.3 + 3.5 +5.7 + ...... + 99.101)
A = 13517400 - 3.171650
A = 13002450
PN
1
27 tháng 10 2016
1.2.3.4+2.3.4.5+3.4.5.6+...+97.98.99.100
4S=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100). 4
4S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4S=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...98.99.100.101-97.98.99.100
4S=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98+99.100+101
4S=98.99.100.101
Vậy S = 98.99.100.101/4 = 24497550
\(\frac{\left(1.2\right)^2}{\left(2.3\right)^2}.\frac{\left(3.4\right)^2}{\left(4.5\right)^2}...\frac{\left(999.1000\right)^2}{\left(1000.1001\right)^2}\)
\(=\frac{1^2.2^2}{2^2.3^2}.\frac{3^2.4^2}{4^2.5^2}...\frac{999^2.1000^2}{1000^2.1001^2}\)
\(=\frac{1^2.2^2.3^2.4^2...999^2.1000^2}{2^2.3^2.4^2.5^2...1000^2.1001^2}\)
\(=\frac{1^2}{1001^2}\)
\(=\frac{1}{1001^2}\)
\(\frac{\left(1.2\right)^2}{\left(2.3\right)^2}.....\frac{\left(999.1000\right)^2}{\left(1000.1001\right)^2}\)
\(=\frac{1^2.2^2}{2^2.3^2}.....\frac{999^2.1000^2}{1000^2.1001^2}\)
\(=\frac{1^2}{3^2}.\frac{3^2}{5^2}.....\frac{999^2}{1001^2}\)
\(=\frac{1^2}{1001^2}=\frac{1}{1002001}\)