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DT
15 tháng 7 2022

\(B=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0.....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =0\)

DT
15 tháng 7 2022

\(B=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{125}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\\ =\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right).\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....0.....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)=0\)

\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\)  \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\)  \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\) \(0\)

24 tháng 7 2018

Các bạn giúp mn với ^^ mn k cho

24 tháng 7 2018

Các bạn giúp mn với ^^ mn k cho

19 tháng 12 2016

Vì dãy số trên có phần tử 1/125 - 1/5^3 = 0 nên tích đó bằng 0

Chúc em học tốt

\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)

\(=0\)

1 tháng 3 2020

thank!haha

1 tháng 3 2020

\(A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\left(\frac{1}{125}-\frac{1}{125}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\cdot0\cdot...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=0\)