已知(a+1)的平方+(2b-3)的平方+|c-1|=0,求ab/3c+a-c/b的值
a+1=0 2b-3=0 c-1=0
a=-1 b=3/2 c=1
ab/3c+a-c/b
=-1*(3/2)/3*1+(-1)-1/(3/2)
=-1/2-1-2/3
=-2 1/6