Question

已知(x+4)(x-2)=0，求(x-1/x-3 - x-4/x)/(x^2-2x)/(x^2+3x)的值

Answer 1

原式=(x²-x-x²+7x-12)/x(x-3)÷x(x-2)/x(x+3)
=(6x-12)/x(x-3)÷(x-2)/(x+3)
=6(x-2)/x(x-3)÷(x-2)/(x+3)
=6(x+3)/x(x-3)

(x+4)(x-2)=0
若x-2=0
则除数x(x-2)/x(x+3)=0，舍去
所以x=-4
所以原式=6(-4+3)/[(-4)(-4-3)=-3/14

Answer 2

原式=(x²-x-x²+7x-12)/x(x-3)÷x(x-2)/x(x+3)
=(6x-12)/x(x-3)÷(x-2)/(x+3)
=(x-2)/x(x-3)÷(x-2)/(x+3)
=(x+3)/x(x-3)

(x+4)(x-2)=0
若x-2=0
则除数x(x-2)/x(x+3)=0，舍去
所以x=-4
所以原式=(-4+3)/[(-4)(-4-3)=-1/28

Answer 3

(x-1/x-3 - x-4/x)/(x^2-2x)/(x^2+3x)
=(x-1/x-3 - x-4/x)]/[x(x-2)]/(x^2+3x)
x≠2，
(x+4)(x-2)=0，x=-4；
=(x-1/x-3 - x-4/x)]/[x(x-2)]/(x^2+3x)
=[(-5/-7)-(-8/-4)/[-4*(-6)]/(32-12)
=-3/1120