Question

已知c/a+c/b=2（a,b,c互不相等且均不为零）求证a-c/(c-b)=a/b

Answer 1

证明：∵c/a+c/b=2
∴[c(1/a+1/b)]=2
[c(a+b)/(ab)]=2
c=2ab/(a+b)
∴(a-c)/(c-b)=[a-2ab/(a+b)]/[2ab/(a+b)-b]
=﹛[a(a+b)-2ab]/(a+b)﹜/﹛[2ab-b(a+b)]/(a+b)﹜
=[(a²+ab-2ab)/(a+b)]/[(2ab-ab-b²)/(a+b)]
=[(a²-ab)/(a+b)]/[(ab-b²)/(a+b)]
=[a(a-b)/(a+b)]/[b(a-b)/(a+b)]
=[a(a-b)/(a+b)]×[(a+b)/b(a-b)]
=a/b
∴(a-c)/(c-b)=a/b, 原命题得证。