(2x^2+ax-1/3×y+1/5)-(1/2×x-2y+1-bx^2)
=(2+b)x^2+(a-1/2)x+(-1/3+2)y-4/5
代数式与x无关,则2+b=0,a-1/2=0
所以,a=1/2,b=-2