Question

数学题求解x平方-6x+y平方+10y+34=0求x平方+3x除以2xy+2y的值

Answer 1

x平方-6x+y平方+10y+34=0

x^2-6x+y^2+10y+34=0,

x^2-6x+9+y^2+10y+25=0,

(x-3)^2+(y+5)^2=0

x-3=0,y+5=0,

x=3,y=-5

所以x平方+3x除以2xy+2y

=(x^2+3x)/(2xy+2y)

=(9+9)/(-30-10)

=18/(-40)

=-9/20