Tính nhanh nhé!:A=(1/125-1/13).(1/125-1/23).(1/125-1/33).......(1/125-1/253)
Bánh mì vô địch!!!!:>
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19 tháng 10 2019
\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(0\)
14 tháng 1 2018
a ) 31 x ( - 125 ) + 225 x 31
= 31 x [ ( - 125 ) + 225 ]
= 31 x 100
= 3100
b ) 26 x ( - 125 ) - 125 x ( - 36 )
= ( - 26 ) x 125 - 125 x ( - 36 )
= 125 x [ ( - 26 ) - ( - 36 ) ]
= 125 x 10
= 1250
14 tháng 1 2018
a) 31 × (-125) + 225 × 31
= 31 x ( -125 + 225 )
= 31 x 100
= 3100
b) 26 × (-125) – 125 × (-36)
= -125 ( 26 - 36 )
= 125x 10
= 1250
d) 17 × (-37) – 23 × 37 – 46 × (-37)
= -629 - 851 + 1702
= 1480
BD
2 tháng 7 2023
a) (4+4)x(2019 +2019 +.....+2019)
= 8 x 125 x 2019
= (8 x 125) x 2019
= 1000 x 2019
= 2019000
b) ( 42+42+...+42)+(125+125+....+125)
= 42 x 125 + 58 x 125
= 125 x ( 42 + 58)
= 125 x 100
= 12500
2 tháng 7 2023
42+42+...+42)+(125+125+....+125)
= 42 x 125 + 58 x 125
= 125 x ( 42 + 58)
= 125 x 100
= 12500
ND
1
1 tháng 8 2019
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
HT
0
ND
0
NH
1
thank!
\(A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\left(\frac{1}{125}-\frac{1}{125}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)\left(\frac{1}{125}-\frac{1}{4^3}\right)\cdot0\cdot...\left(\frac{1}{125}-\frac{1}{25^3}\right)\\ A=0\)