Question

[nguyen tuan]

1999*(2000^2=2001)-2001(2000^2-1999)

Answer 1

[Nguyễn Lê Phước Thịnh]

Sửa đề: \(1999\left(2000^2+2001\right)-2001\left(2000^2-1999\right)\)

Đặt A=\(1999\left(2000^2+2001\right)-2001\left(2000^2-1999\right)\)

Đặt x=1999

=>x+1=2000; x+2=2001

=>\(A=x\left[\left(x+1\right)^2+x+2\right]-\left(x+2\right)\left[\left(x+1\right)^2-x\right]\)

\(=x\left(x^2+2x+1+x+2\right)-\left(x+2\right)\left(x^2+2x+1-x\right)\)

\(=x\left(x^2+3x+3\right)-\left(x+2\right)\left(x^2+x+1\right)\)

\(=x^3+3x^2+3x-x^3-x^2-x-2x^2-2x-2\)

\(=-2\)

Answer 2

[Nguyễn Thị Thương Hoài]

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