[(x-2)/(x^2+2x)-(x-1)(x^2+4x+4)]/[(x^2-5x+4)/(x^4+2x^3+8x+16)]
={(x-4)/[x(x+2)^2]}/{(x-1)(x-4)/[(x+2)^2*(x^2-2x+4)]}
=(x^2-2x+4)/(x^2-x)
分子=(x-2)/[x(x+2)]-(x-1)/(x+2)^2=[(x-2)(x+2)-(x-1)x]/[x(x+2)^2]=(x-4)/[x(x+2)^2] x^4+2x^3+8x+16=x^3(x+2)+8(x+2)=(x+2)[x^3+8]=(x+2)(x+2)(x^2-2x+4) 分母=(x-4)(x-1)/[(x+2)^2(x^2-2x+4)] 因此原分式=(x-4)/[x(x+2)^2]*[(x+2)^2(x^2-2x+4)]/[x-4)(x-1)]=(x^2-2x+4)/[x(x-1)]
答：原式=（X-Y)²-1/X-Y-1 令T=X-Y 原式=T²-1/T-1=T＋1 带入T=X-Y 原式=X-Y＋1详情>>
答：详情>>
