求值：[(x+2y-3/2)(x-2y+3/2)+2y(2y-3)+(2/3)⁻²]÷（-3x)⁻¹其中x=-1,y=2007/2006.
原式={[x+(2y-3/2)][x-(2y-3/2)]+2y(2y-3)+9/4}×(-3x)
={[x²-(2y-3/2)²]+2y(2y-3)+9/4}×(-3x)
=[x²-4y²+6y-9/4+4y²-6y+9/4]×(-3x)
=x²(-3x)=-3x³=-3×(-1)=3