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a) ( 1000-13) . ( 1000-23) . ( 1000-33) ...( 1000 -503)
\(=\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-10^3\right)\cdot.....\cdot\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\cdot\left(100-2^3\right)\cdot...\cdot\left(1000-1000\right)\cdot...\cdot\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot......\cdot0\cdot......\left(1000-50^3\right)\)
\(=0\)
b) (1/125-1/13) . (1/125-1/23).( 1/125-1/33)...( 1/125-1/253)
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{5^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{125}\right)\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\cdot\left(\frac{1}{125}-\frac{1}{2^3}\right)\cdot....\cdot0\cdot...\cdot\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
\(\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)....\left(1000-50 ^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)......\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)......\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)....0.....\left(1000-50^3\right)\)
\(=0\)
Trong tích B có thừa số \(1000-10^3=1000-1000=0\)
Vậy B = 0
\(\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-10^3\right)\cdot,,,\cdot\left(1000-50^3\right)\)
Vì \(1000-10^3=1000-1000=0\) nên \(\left(1000-1^3\right)\cdot\left(1000-2^3\right)\cdot...\cdot\left(1000-50^3\right)=0\)
(1000-1^3)(1000-2^3)...(1000-10^3)...(1000-50^3)=(1000-1^3).(1000-2^3)...0...(1000-50^3)=0
Bạn xinh xinh gì đó ơi, tk mình nhé