若代数式（2x^2+ax-y+6)-(2bx^2-2x+5y-1)的值与字母x的取值无关,

（2x^2+ax-y+6)-(2bx^2-2x+5y-1)
=2x^2+ax-y+6-2bx^2+2x-5y+1
=(2-2b)x^2+(a+2)x-6y+7
和x无关则系数为0
2-2b=0,a+2=0
a=-2,b=1
所以原式=1/3a³-2b²-1/4a³+3b²
=1/12a³+b²
=-2/3+1
=1/3谢咯！不客气