Question

求证:无论x,y为何有理数,多项式x^2+y^2-2x+6y+16的值恒为正数.

Answer 1

x^2+y^2-2x+6y+16

=x^2-2x+y^2+6y+16

=x^2-2x+1+y^2+6y+9+6

=(x-1)^2+(y+3)^2+6

因为(x-1)^2>=0,(y+3)^2>=0

所以(x-1)^2+(y+3)^2+6>0

即无论x,y为何有理数,多项式x^2+y^2-2x+6y+16的值恒为正数