Can someone help me solve this maths problem? Need to simplify. Thks.?

Question

(1/x^2-3x) - (x-1)/(x^2-9)

Answer 1

1/(x^2-3x) - (x-1)/(x^2-9)
= 1/[(x)(x-3)] - (x-1)/[(x-3)(x+3)]
= [(x+3)-(x-1)(x)] / [(x)(x-3)(x+3)]
= [(x+3)-(x^2-x)] / [(x)(x-3)(x+3)]
= (2x+3-x^2) / [(x)(x-3)(x+3)]
= -(x^2-2x-3) / [(x)(x-3)(x+3)]
= -[(x-3)(x+1)] / [(x)(x-3)(x+3)]
= - (x+1) / [ (x)(x+3)]

Answer 2

I am assuming your expression is wrong. Instead I am going to simplify this

I/(x^2-3x) - (x-1)/(x^2-9)
= 1/(x(x-3) - (x-1)/(x+3)(x-3)

LCM for x(x-3) and (x+3)(x-3) = x(x-3)(x+3)

(x+3)/x(x+3)(x-3) - x(x-1)/x(x+3)(x-3)

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