解析：已知：a=(1/5)x+20,b=(1/5)x+19,c=(1/5)x+21,那么：a-b=1,a-c=-1,b-c=-2所以：a^2+b^2+c^2-ab-bc-ca=(1/2)×(2a²+2b²+2c²-2ab-2bc-2ca)=(1/2)×[(a-b)²+(a-c)²+(b-c)²]=(1/2)×...