Question

A^2+4AB+5B^2-2B+1=0,则A-B=

Answer 1

原式=A^2+4AB+4B^2+B^2-2B+1=0

(A^2+4AB+4B^2)+(B^2-2B+1)=0

(A+2B)^2+(B-1)^2=0

因为(A+2B)^2>=0,(B-1)^2>=0,(A+2B)^2+(B-1)^2=0

所以A+2B=0,B-1=0

A=-2,B=1

A-B=-2-1=-3