已知abc≠0,且a/b=b/c=c/a,则3a+2b+c/a−2c−3c=_.
问题描述:
已知abc≠0,且
=a b
=b c
,则c a
=______. 3a+2b+c a−2c−3c
答
设
=a b
=b c
=t(t≠0),则c a
a=bt①
b=ct②
c=at③
将①②③相乘得abc=abct3,
∵abc≠0,∴t=1,∴a=b=c,
∴
=3a+2b+c a−2c−3c
=-6a −4a
,3 2
故答案为-
.3 2