已知(a+1)2-(3a2+4ab+4b2+2)=0,求a,b的值.
∵(a+1)2-(3a2+4ab+4b2+2)=0,
∴2a2-2a+4b2+4ab+1=0,
∴(a-1)2+(a+2b)2=0,
∴a-1=0,a+2b=0,
解得a=1,b=-
.1 2
故a=1,b=-
.1 2
已知(a+1)2-(3a2+4ab+4b2+2)=0,求a,b的值.
∵(a+1)2-(3a2+4ab+4b2+2)=0,
∴2a2-2a+4b2+4ab+1=0,
∴(a-1)2+(a+2b)2=0,
∴a-1=0,a+2b=0,
解得a=1,b=-
.1 2
故a=1,b=-
.1 2