已知a/2=b/3=c/4,求(a+2b+3c)/(a-2b-3c)
问题描述：
已知a/2=b/3=c/4,求(a+2b+3c)/(a-2b-3c)
答
a/2=b/3=c/4 ---- 6a=4b=3c
a+2b+3c= a+ 3a + 6a = 10a
a-2b-3c = a - 3a -6a = -8a
结果-5/4
答
设a/2=b/3=c/4=t
所以a=2t，b=3t，c=4t代入到代数式得
原式=（2t+2*3t+3*4t)/(2t-2*3t-3*4t)
=20t/-16t
=-5/4
答
a/2=b/3=c/4
b=3/2a
c=2a
(a+2b+3c)/(a-2b-3c)
=(a+3a+6a)/(a-3a-6a)
=10a/-8a
=-5/4