(a+1)^2=?(a-1)^2=?a^2+b^2+c^2-2a+4b-6c+14=0,求a+b+c的值是?

题目

(a+1)^2=?(a-1)^2=?a^2+b^2+c^2-2a+4b-6c+14=0,求a+b+c的值是?

答案

a^2+b^2+c^2-2a+4b-6c+14=0==>
(A-1)^2+(B+2)^2+(C-3)^2=0==>必须A-1=0 B+2=0 C-3=0
A=1,B=-2,C=3
所以A+B+C=1-2+3=2