Question

若a2+b2-6a+2b+10=0 ,求a2+b2-ab

Answer 1

a2+b2-6a+2b+10=0

a^2-6a+9+b^2+2b+1=0

(a-3)^2+(b+1)^2=1

因为(a-3)^2>=0 (b+1)^2>=0

所以(a-3)^2=(b+1)^2=0

a-3=b+1=0

a=3 b=-1

a^2+b^2-ab

=3^2+(-1)^2-3*(-1)

=9+1+3

=13