Question

(x+2/2x^2-3x-x^2+2x/2x^2+x-6)/（1/2-1/4x-6）

Answer 1

{ (x+2)/(2x^2-3x) - (x^2+2x)/(2x^2+x-6) } / {1/2-1/(4x-6) }
= { (x+2)/[x(2x-3)] - x(x+2)/[(x+2)(2x-3)]} / {1/2-1/[2(2x-3)] }
= { (x+2)/[x(2x-3)] - x/(2x-3) } / {1/2-1/[2(2x-3)] }
分子分母同乘以2x(2x-3)：
= {2(x+2) - 2x^2} / {x(2x-3)-x }
= {2x+4- 2x^2} / {2x^2-3x-x }
=- {2x^2-2x+4} / {2x^2-4x }
=- 2(x+1)(x-2) / {2x(x-2) }
= - (x+1) / x

Answer 2

(x+2/2x^2-3x-x^2+2x/2x^2+x-6)/（1/2-1/4x-6）
=[(x+2)/x(2x-3)-x(x+2)/(2x-3)(x+2)]/[1/2-1/2(2x-3)]
=[(x+2)/x(2x-3)-x/(2x-3)]/[(2x-3-1)/2(2x-3)]
=[(x+2-x²)/x(2x-3)]*2(2x-3)/2(x-2)]
=-[(x²-x-2)/x]*[1/(x-2)]
=-(x-2)(x+1)/x(x-2)
=-(x+1)/x