（x^3+4）/（x^4+8）=(x³-3x²+x+3x²-x+4)/[(x²)²+8]=[x(x²-3x+1)+3x²-9x+3+8x+1]/[(3x-1)²+8]=[3(x²-3x+1)+8x+1]/(9x²-6x+1+9)=(8x+1)/(9x²-27x+9+21x+1)=(...然后拿（x^3+4）/（x^4+8）=(x³-3x²+x+3x²-x+4)/[(x²)²+8]=[x(x²-3x+1)+3x²-9x+3+8x+1]/[(3x-1)²+8]=[3(x²-3x+1)+8x+1]/(9x²-6x+1+8)=(8x+1)/(9x²-27x+9+21x)=(8x+1)/(21x) 我也没有办法了。我也算到这里了，你这叫我咋办啊（x^3+4）/（x^4+8）=(x³+1+3)/(x^4-1+9)=[(x+1)(x²-x+1)+3]/[(x²+1)(x²-1)+9]=[(x+1)(x²-3x+1+2x)+3]/[3x*(x²-3x+1+3x-2)+9]=(2x²+2x+3)/(9x²-6x+9)=(2x²-6x+2+8x+1)/(9x²-27x+9+21x)=(8x+1)/(21x)