\(M=\frac{1}{125\cdot2^3}\cdot\frac{1}{125\cdot3^3}\cdot\frac{1}{125\cdot4^3}\cdot....\cdot\frac{1}{125\cdot20^3}\)
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23 tháng 12 2016
A=1/1x2 +1/2x3 +... +1/18x19 + 1/19x20
Nhận xét 1/1x2 = 1/1 -1/2 ; 1/2x3=1/2-1/3; ... ;1/18x19=1/18-1/19 ; 1/19x20=1/19-1/20
ta có A=1/1 - 1/2 + +1/2 -1/3+1/3- +1/18-1/19+1/19-1/20
A=1/1 - 1/20
A=20/20 - 1/20
A=(20-1)/20
A=19/20
Vậy A=19/20
23 tháng 12 2016
A =\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ...+\(\frac{1}{18.19}\)+\(\frac{1}{19.20}\)
A = 1 - \(\frac{1}{2}\)+\(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+....+\(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)
A = 1 - \(\frac{1}{20}\)( Vì đã triệt tiêu )
A = \(\frac{19}{20}\)
1 tháng 8 2019
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
T
13 tháng 7 2019
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
19 tháng 10 2019
\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(0\)
6 tháng 8 2018
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
MN
27 tháng 2 2016
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}=\frac{1^2.2^2.3^2...9^2}{1.2.2.3.3.4.4...9.10}=\frac{1.2^2.3^2...9^2}{1.2^2.3^2.4^2...10^2}=\frac{1}{10^2}=\frac{1}{100}\)
Ta có \(M=\frac{1}{125\cdot2^3}\cdot\frac{1}{125\cdot3^3}\cdot\frac{1}{125\cdot4^3}\cdot...\cdot\frac{1}{125\cdot20^3}\)
\(\Rightarrow M=0\)