Question

已知a²+b²+c²-10a-24b-26c+338=0,求a,b,c

Answer 1

即(a²-10a+25)+(b²-24b+144)+(c²-26c+169)=0

(a-5)²+(b-12)²+(c-13)²=0

所以a-5=0,b-12=0,c-13=0

所以a=5,b=12,c=13