初二奥数题两道。。。

Question

拜托了

1。已知：（a＋b＋c）（ab＋bc＋ca）＝abc
求证a＝－b或b＝－c或c＝－a

2。若(a+2b)/5=(3b-c)/3=(2c-a)/7,
求(c-2b)/(3a+2b)

Answer 1

1.(a+b+c)(ab+bc+ca)-abc=0
=a^2b+2abc+ca^2+ab^2+b^2c+bc^2+c^2a
=(a^2b+ab^2)+(bc^2+ac^2)+(2cab+ca^2+cb^2)
=ab(a+b)+c^2(a+b)+c(a+b)^2
=(a+b)(ab+c^2+ac+cb)
=(a+b)[(ab+bc)+(c^2+ac)]
=(a+b)(a+c)(b+c) =0
所以a＝－b或b＝－c或c＝－a

2.假设 (a+2b)/5=(3b-c)/3=(2c-a)/7 = t
a = 5t - 2b
c = 3b - 3t
带入 2c - a = 7t, 6b - 6t - 5t +2b = 7t, 8b = 18t, b = 9t/4
如果 t = 0, b = 0 = a = c,分母 3a+2b = 0, 分式无意义
如果 t <> 0
(c-2b)/(3a+2b) = (3b - 3t - 2b)/(15t - 6b + 2b)
= (b - 3t ) / (15t - 4b)
= (9t/4 - 3t) / (15t - 9t)
= (9 -12) / 24
= -3 / 24
= -1/8

Answer 2

1. （a＋b＋c）（ab＋bc＋ca）＝abc
全部展开化成(a+b)(a+c)(b+c)=0
2. a+2b=5k
3b-c=3k
2c-a=7k
a=1/2k, b=9k/4, c=15k/4

(c-2b)/(3a+2b) =-1/8