Question

(a+b)/(a-b)=(b+c)/2(b-c)=(c+a)/3(c-a),abc互不相等,证8a+9b+5c=o

Answer 1

不妨设(a+b)/(a-b)=(b+c)/2(b-c)=(c+a)/3(c-a)=k,则

a+b=k(a-b)

b+c=2k(b-c)

c+a=3k(c-a)

三式相加得a+b+c=k(-a+b/2+c/2)与三式相减得

a=k(-a-3b/2+5c/2)

b=k(2a+b/2-5c/2)

c=k(-2a+3b/2+c/2)

所以3a+9b+5c=k*(0*a+0*b+0*c)=0

证毕!