Question

\\(\dfrac{x + 3}{x^2 - 2x + 1} + \dfrac{x}{x^2 - 3x + 2}\\)

\\(\dfrac{x}{x^2 - 4x + 4} - \dfrac{2}{x^2 - 4}\\)

Explanation:

Step1: Factor all denominators

First fraction numerator: $x+3$
First fraction denominator: $x^2-2x+1=(x-1)^2$
Second fraction numerator: $x$
Second fraction denominator: $x^2-3x+2=(x-1)(x-2)$
Third fraction numerator: $x$
Third fraction denominator: $x^2-4x+4=(x-2)^2$
Fourth fraction numerator: $2$
Fourth fraction denominator: $x^2-4=(x-2)(x+2)$

Step2: Simplify numerator of the complex fraction

Find common denominator $(x-1)^2(x-2)$:

$$\begin{align*} &\frac{x+3}{(x-1)^2} + \frac{x}{(x-1)(x-2)}\\ =&\frac{(x+3)(x-2) + x(x-1)}{(x-1)^2(x-2)}\\ =&\frac{x^2+x-6 + x^2-x}{(x-1)^2(x-2)}\\ =&\frac{2x^2-6}{(x-1)^2(x-2)}\\ =&\frac{2(x^2-3)}{(x-1)^2(x-2)} \end{align*}$$

Step3: Simplify denominator of the complex fraction

Find common denominator $(x-2)^2(x+2)$:

$$\begin{align*} &\frac{x}{(x-2)^2} - \frac{2}{(x-2)(x+2)}\\ =&\frac{x(x+2) - 2(x-2)}{(x-2)^2(x+2)}\\ =&\frac{x^2+2x - 2x+4}{(x-2)^2(x+2)}\\ =&\frac{x^2+4}{(x-2)^2(x+2)} \end{align*}$$

Step4: Rewrite complex fraction as multiplication

Divide by a fraction = multiply by its reciprocal:

$$ \frac{\frac{2(x^2-3)}{(x-1)^2(x-2)}}{\frac{x^2+4}{(x-2)^2(x+2)}} = \frac{2(x^2-3)}{(x-1)^2(x-2)} \times \frac{(x-2)^2(x+2)}{x^2+4} $$

Step5: Cancel common factors

Cancel $(x-2)$ from numerator and denominator:

$$ \frac{2(x^2-3)(x-2)(x+2)}{(x-1)^2(x^2+4)} $$

Note: $x
eq1, x
eq2, x
eq-2$ to avoid division by zero.

Answer:

$\frac{2(x^2-3)(x^2-4)}{(x-1)^2(x^2+4)}$ or expanded form $\frac{2(x^4-7x^2+12)}{(x-1)^2(x^2+4)}$ (where $x
eq1,2,-2$)