Question

abc是整数,若a平方+b平方+c平方+4a-8b-14c+69=0,求a+2b-3c的值

Answer 1

a平方+b平方+c平方+4a-8b-14c+69=a^2+4a+4+b^2-8b+16+c^2-14c+49

=(a+2)^2+(b-4)^2+(c-7)^2=0,因为a+2)^2,(b-4)^2,(c-7)^2都大于等于0,所以a+2=0,b-4=0,c-7=0,所以 a=-2,b=4,c=7,所以a+2b-3c=-2+8-21=-11