Question

若实数x,y满足x^2+3y^2-12y+12=0,则y^x的值是?

Answer 1

x^2+3y^2-12y+12=x^2+3(y^2-4y+4)

=x^2+3(y-2)^2=0, 因为x^2≥0, 3(y-2)^2≥0

所以: x=0, y-2=0,

y^x=2^0=1