若a,b,c为两两不等的有理数,求证：√1\(a-b)^2+1\(b-c)^2+1\(c-a)^2为有理数.

Question

若a,b,c为两两不等的有理数,求证：√1\(a-b)^2+1\(b-c)^2+1\(c-a)^2为有理数.

Answer 1

证明：令x=a-b y=b-c
则c-a=(b-y)-(x+b)=-x-y
原式=√[1/x^2+1/y^2+1/(-x-y)^2]
=√{[y^2(x+y)^2+x^2(x+y)^2+x^2y^2]/[x^2y^2(x+y)^2]}
=√{[(x+y)^2(x^2+y^2)+x^2y^2]/[x^2y^2(x+y)^2]}
=√{[(x+y)^2-xy]^2/[x^2y^2(x+y)^2]}
=|[(x+y)^2-xy]/[xy(x+y)]|
因为x和y是有理数,所以|[(x+y)^2-xy]/[xy(x+y)]|也是有理数
即原式是有理数