我们来看一道题 :已知a、b、c为两两互不相等的有理数 .求证1(a -b) 2 + 1(b -c) 2 + 1(c -a) 2为有理数 .为了运算的简化 ,我们不妨设a bc,且设a=b +m ,c=b-n(m 0 ,n0 ) ,则a-b=m ,b -c=n ,c-a =-m-n ,∴ 1(a-b) 2 + 1(b-c) 2 + 1(c-a) 2=1m2 + 1n2 + 1(m +n) 2=n2 (m+n) Let’s look at a problem: we know that a, b, and c are rational numbers that are not equal to each other. Proof 1(a -b) 2 + 1(b -c) 2 + 1(c -a) 2 is a rational number. The simplification of the operation, we may wish to set a bc, and let a = b + m, c = bn (m 0, n0), then ab = m, b - c = n, ca = - mn, ∴ 1 (ab) 2 + 1(bc) 2 + 1(ca) 2=1m2 + 1n2 + 1(m +n) 2=n2 (m+n)