Question

Tính giá trị của biểu thức:

B=\(\text{x}^{\text{3}}+3\text{x}^{\text{2}}+3\text{x}^{\text{2}}y+3x\text{y}^{\text{2}}+\text{y}^{\text{3}}+3\text{y}^{\text{2}}+6xy+3x+3y+2019 \) Biết x+y=101

Answer 1

\(B=x^3+3x^2+3x^2y+3xy^2+y^3+3y^2+6xy+3x+3y+2019\)

\(=\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+2019\)

\(=\left[\left(x+y\right)^3-3\left(x+y\right)^2+3\left(x+y\right)+1\right]+2018\)

\(=\left(x+y-1\right)^3+2018\)

Mà \(x+y=101\)

\(B=\left(101-1\right)^3+2018=1002018\)