Question

先化简再求值(x^2-4/x^2+x+1)^2/[x^3-3x^2+2x/(x-1)(x^2+x+1)],其中x=1/3

打错了
第一个分母是x^2+x+1

Answer 1

原式=[(x+2)²(x-2)²/(x²+x+1)²]*[(x-1)²(x²+x+1)²/x²(x-1)²(x-2)²]*[x³/(x+2)³]
=x³(x+2)²(x-2)²(x-1)²(x²+x+1)²/[x²(x²+x+1)²(x-1)²(x-2)²(x+2)³]
=x/(x+2)
x=1/3,则x/(x+2)=1/7

追问

谢谢

Answer 2

原式=(x+2)(x-2)/(x²+x+1)²*(x-1)(x²+x+1)/x(x²-3x+2)
=(x+2)(x-2)/(x²+x+1)²*(x-1)(x²+x+1)/x(x-1)(x-2)
=(x+2)(x-2)/(x²+x+1)²*(x²+x+1)/x(x-2)
=(x+2)/[x(x²+x+1)]
=(1/3+2)/[1/3*(1/9+1/3+1)]
=63/13