Answer: [tex]\bold{\dfrac{x}{x^2-3x+2}+\dfrac{1}{x^2-3x+2}}[/tex]
Step-by-step explanation:
[tex]\dfrac{x^2-3x-2}{(x-3)(x-2)(x-1)} \qquad ;x\neq{1, 2, 3}\\\\\\\\\text{Factor the numerator}\quad \longrightarrow \quad \dfrac{(x-3)(x+1)}{(x-3)(x-2)(x-1)}\\\\\\\text{Cancel out (x-3)}\quad \longrightarrow \quad \dfrac{x+1}{(x-2)(x-1)}\\\\\\\text{To separate them, just place each term in the numerator}\\ \text{over the entire denominator.}\\\\\dfrac{x}{(x-2)(x-1)}+\dfrac{1}{(x-2)(x-1)}\\\\\\=\large\boxed{\dfrac{x}{x^2-3x+2}+\dfrac{1}{x^2-3x+2}}[/tex]
