Question

已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2)

Answer 1

即(x^2-8x+16)+(y^2+12y+36)=0

(x-4)^2+(y+6)^2=0

所以x-4=0,y+6=0

x=4,y=-6

所以原式=1/2x-1/(x-y)*[(x-y)/x-(x+y)(x-y)]

=1/2x-1/x+(x+y)

=-1/(2x)+x+y

=-1/8+4-6

=-17/8