Question

已知x+y=8,xy=12,求(1)x2y+xy2(2)x2-xy+y2(3)x-y

Answer 1

(1)

x^2y+xy^2

=xy(x+y)

=12*8

=96

(2)

x^2-xy+y^2

=x^2+2xy+y^2-3xy

=(x+y)^2-3xy

=8^2-3*12

=64-36

=28

(3)

(x-y)^2

=x^2-2xy+y^2

=x^2+2xy+y^2-4xy

=(x+y)^2-4xy

=8^2-4*12

=64-48

=16

x-y=±4