Question

[Minh Hoang Hai]

Tính nhanh :

a) \(A=\dfrac{2004^3+1}{2004^2-2003}\)

b) \(B=\dfrac{2005^3-1}{2005^2+2006}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)

b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)