Question

已知x2-4x-y2-10+29=0,求：x²y²+2x³y²+x^4y²的值

Answer 1

应该是x^2-4x+y^2-10y+29=0吧

x^2-4x+y^2-10y+29=0

(x^2-4x+4)+(y^2-10y+25)=0

(x-2)^2+(y-5)^2=0

x=2,y=5

代入得

x^2y^2+2x^3y^2+x^4y^2

=x^2y^2(1+2x+x^2)

=x^2y^2(1+x)^2

=4*25*9

=900