Question

已知x²+y²-4x-2y+5=0,求log(x)y^x的值

Answer 1

已知x2+y2-4x-2y+5=0,求log(x)y^x的值.

x^2-4x+4+y^2-2y+1=0

(x-2)^2+(y-1)^2=0

x-2=0 y-1=0

x=2 y=1

log(x)y^x=log(2)1^2=log(2)1=0