Question

Tính giá trị biểu thức

N=\(x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)

Biết x + y = 1

#Giúp_nhoa

#Ken

Answer 1

\(N=x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)

\(N=x^3+y^3+6x^2y^2+3xy\left[\left(x+y\right)^2-2xy\right]\)

\(N=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2+3xy-6x^2y^2\)

\(N=x^2-xy+y^2+3xy\)

\(N=\left(x+y\right)^2\)

\(N=1\)

Answer 2

\(x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]\)

\(=x^2-xy+y^2+6x^2y^2+3xy-6x^2y^2\)(  Do  \(x+y=1\))

\(=\left(x+y\right)^2-2xy-xy+3xy+6x^2y^2-6x^2y^3\)

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