化简求值(x²-4／x²-x-1)²÷(x³-3x²+2x／x³-1)²（x／x+2)³其中x=

题目，应该 是这样的吧
(x²-4／x²-x+1)²÷(x³-3x²+2x／x³-1)²（x／x+2)³ 其中x= -2/3
原式＝{[(x+2)(x-2)/(x^2-x+1)]*[(x-1)(x^2+x+1)/x(x-)(x-2)]}^2*（x／x+2)³
=[(x+2)/(x-2)]^2*（x／x+2)³
=x^3/(x+2)(x-2)
当x=-2/3时，原式＝1/12

(x²-4／x²-x-1)²÷(x³-3x²+2x／x³-1)²（x／x+2)³
=[(x+2)(x-2)/(x^2-x-1)]^2*[(x-1)(x^2+x+1)/x(x-2)(x-1)]^2*[x^3/(x+2)^3]
=[(x+2)^2(x-2)^2/(x^-x-1)^2]*[(x^2+x+1)^2/x^2(x-2)^2]*[x^3/(x+2)^3
=x(x^2-x-1)^2/(x+2)(x^2+x+1)^2=[x/(x+2)]*[(x^2-x-1)/(x^2+x+1)]^2
代x=-2/3
得原式=[(-2/3)/(2-2/3)]*[(1/9)/(19/9)]^2
=(-1/4)*(1/19)^2=1/1444

题写的额部清楚，特别是除号所针对的分母分子，不清楚