\(\dfrac{x-14}{x^2-4x}\) - \(\dfrac{3}{2x}\) + \(\dfrac{x+1}{2x-8}\) Đk x #0; x # 4
= \(\dfrac{x-14}{x(x-4)}\) - \(\dfrac{3}{2x}\) + \(\dfrac{x+1}{2(x-4)}\)
= \(\dfrac{2.(x-14)-3.(x-4)+x.(x+1)}{2.x.(x-4)}\)
= \(\dfrac{2x-28-3x+12+x^2+x}{2x(x-4)}\)
= \(\dfrac{x^2-16}{2x(x-4)}\)
= \(\dfrac{(x-4).(x+4)}{2x(x-4)}\)
= \(\dfrac{x+4}{2x}\)