Question

已知(a+1)平方-(3a的平方+4ab+4b+2)=0求a.b的值

Answer 1

（a+1）^2-(3a^2+4ab+4b^2+2)

=a^2+2a+1-3a^2-4ab-4b^2-2

=-2a^2+2a-1-4ab-4b^2=0

所以2a^2-2a+1+4ab+4b^2=0

(a^2-2a+1)+(a^2+4ab+4b^2)=0

(a-1)^2+(a+2b)^2=0

因为(a-1)^2>=0,(a+2b)^2>=0

所以a-1=0,a+2b=0

a=1,b=-1/2