若abc是有理数,且有/a/是1,/b-2/+（3a+2c）的平方=0,试求abc除a+b+c的值

|a|=1,那么a=±1
|b-2|+(3a+2c)²=0
那么b-2=0,b=2
3a+2c=0,c=-3/2或c=3/2
(1)a=1,b=2,c=-3/2
（a+b+c)/abc
=(1+2-3/2)/[1×2×(-3/2)]
=(-3/2)/(-3)
=3/2×1/3
=1/2
(2)a=-1,b=2,c=3/2
(a+b+c)/abc
=(-1+2+3/2)/[(-1)×2×3/2]
=(5/2)/(-3)
=-5/6