已知a与b互为相反数,c与d互为倒数,x=3（a-1)-(a-2b),y=c^2d+d^2-((d/c)+c-1),求(2x+y)/3-(3x-2y)/6的值

a+b=0
cd=1
则c²d=(cd)c=1*c=c
d=1/c
所以d/c=*d=d²
所以x=3a-3-a+2b
=2a+2b-3
=2(a+b)-3
=0-3
=-3
y=c+d²-d²-c+1=1
原式=[2(2x+y)-(3x-2y)]/6
=(4x+2y-3x+2y)/6
=(x+4y)/6
=(-3+4)/6
=1/6