已知a²+b²+c²-ab-3b-2c+4=0,那么：a²-ab+(b²/4) +(3b²/4) -3b+3+c²-2c+1=0即：(a- b/2)²+ 3(b/2 - 1)²+(c-1)²=0则有：a - b/2=0,b/2 -1=0,c-1=0解得：a=b/2=1,b=2,c=1所以：a+b+c=1+2+1=4
