Question

a/b/c是不相等实数,求证：(b-c)/(a-b)(a-c)+(c-a)/(b-c)(b-a)+(a-b)/(c-a)

Answer 1

证明：因为

(b-c)/[(a-b)(a-c)]=[(b-a)+(a-c)]/[(a-b)(a-c)]=1/(a-b)+1/(c-a),

(c-a)/[(b-c)(b-a)]=[(c-b)+(b-a)]/[(b-c)(b-a)]=1/(b-c)+1/(a-b),

(a-b)/[(c-a)(c-b)]=[(a-c)+(c-b)]/[(c-a)(c-b)]=1/(b-c)+1/(c-a),

三式相加得(b-c)/[(a-b)(a-c)]+(c-a)/[(b-c)(b-a)]+(a-b)/[(c-a)(c-b)]=2/(a-b)+2/(b-c)+2/(c-a),

命题得证.