已知a2＝b3＝c5，且a+b+c≠0，则2a+3b−2ca+b+c=310310．

Question

已知(a)/(2)＝(b)/(3)＝(c)/(5)，且a+b+c≠0，则(2a+3b−2c)/(a+b+c)=

(3)/(10)

(3)/(10)

．

Answer 1

设(a)/(2)=(b)/(3)=(c)/(5)=k（k≠0），
则a=2k，b=3k，c=5k，
所以，(2a+3b−2c)/(a+b+c)=(2×2k+3×3k−2×5k)/(2k+3k+5k)=(3)/(10)．
故答案为：(3)/(10)．