\(A=\left(1000-1^3\right)\left(1000-2^3\right)......\left(1000-25^3\right)\)
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NN
21 tháng 8 2020
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
BD
14 tháng 12 2016
Vì 103 = 1000 nên :
( 1000 - 103 ) = 0
Số nào nhân với 0 cũng bằng 0
Vậy A = 0
BT
27 tháng 3 2018
\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)
=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)
=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)
Đáp số: C=1
4 tháng 2 2017
Vì trong dãy trên sẽ có 1000-10\(^3\)=0
\(\Rightarrow\)(1000-1)(1000-2\(^3\))...(1000-50\(^3\))=0
4 tháng 2 2017
Tính: (1000−13).(1000−23).(1000−33)......(1000−503)=..........
Ta có : 1000 - 13 = 1000 - 1000 = 0
Nên : (1000−13).(1000−23).(1000−33)......(1000−503)= 0
Vậy ...
PT
23 tháng 2 2017
\(\left(1000-1^3\right).\left(1000-2^3\right).\left(1000-3^3\right)....\left(1000-50^3\right)\)
\(=\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-10^3\right)....\left(1000-50^3\right)\)
\(=\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-1000\right)....\left(1000-50^3\right)\)
\(=\left(1000-1^3\right).\left(1000-2^3\right)...0...\left(1000-50^3\right)\)
\(=0\)
\(A=\left(1000-1^3\right)\left(1000-2^3\right).............\left(1000-25^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right).........\left(1000-10^3\right).........\left(100-25^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)..........\left(1000-1000\right).......\left(1000-25^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right).....0........\left(1000-25^3\right)\)
\(=0\)
\(A=\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-25^3\right)\)
\(=\left(10^3-1^3\right)\left(10^3-2^3\right)...\left(10^3-25^3\right)\)
\(=\left(10^3-1^3\right)\left(10^3-2^3\right)...\left(10^3-10^3\right)...\left(10^3-25^3\right)\)
\(=0\)
Vậy A = 0.