Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
Vì trong dãy trên sẽ có 1000-10\(^3\)=0
\(\Rightarrow\)(1000-1)(1000-2\(^3\))...(1000-50\(^3\))=0
Tính:
Ta có : 1000 - 13 = 1000 - 1000 = 0
Nên : (= 0
Vậy ...
\(\left(1000-1^3\right).\left(1000-2^3\right).\left(1000-3^3\right)....\left(1000-50^3\right)\)
\(=\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-10^3\right)....\left(1000-50^3\right)\)
\(=\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-1000\right)....\left(1000-50^3\right)\)
\(=\left(1000-1^3\right).\left(1000-2^3\right)...0...\left(1000-50^3\right)\)
\(=0\)
Vì 103 = 1000 nên :
( 1000 - 103 ) = 0
Số nào nhân với 0 cũng bằng 0
Vậy A = 0
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
Trong biểu thức trên có chứa (1000-103), mà (1000-103)=1000-1000=0
Do đó tích trên bằng 0
\(\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)...\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-1000\right)...\left(1000-50^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\cdot0\cdot\left(1000-50^3\right)\)
\(=0\)
