若a=kx+1,b=kx-3,c=4-2kx,求a2+b2+c2+2ab+2bc+2ac

解：a2+b2+c2+2ab+2bc+2ac
=(a2+2ab+b2)+2(a+b)c+c2
=(a+b)2+2(a+b)c+c2
=(a+b+c)2
代入：a=kx+1,b=kx-3,c=4-2kx
得：（kx+1+kx-3+4-2kx)2
=4

若a=kx+1,b=kx-3,c=4-2kx
那么,a-b=4
b+a=2kx-2=(4-c)-2=-c+2
所以,a+b+c=2
而(a+b+c)^2=a^2+b^2+c^+2ab+2ac+2bc=4