Question

P=-3xy(xy-2y^2)-x^2(x^2-y^2)+2y^2(x^2-3xy tại x=-2 y=2011 tại x=-2;y=2011

Answer 1

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

\(P=-3x^2y^2+6xy^3-x^4+x^2y^2+2x^2y^2-6xy^3\)

\(P=-x^4\)

Thay x = -2 vào P, ta có:

\(P=-\left(-2\right)^4=-16\)

Answer 2

Ta có: \(P=-3xy\left(xy-2y^2\right)-x^2\left(x^2-y^2\right)+2y^2\left(x^2-3xy\right)\)

\(=-3x^2y^2+6xy^3-x^4+x^2y^2+2x^2y^2-6xy^3\)

\(=-x^4\)

\(=-16\)