Question

\frac{x + 3}{x^2 - 2x + 1} + \frac{x}{x^2 - 3x + 2}

\frac{x}{x^2 - 4x + 4} - \frac{2}{x^2 - 4}

Explanation:

Step1: Factor all denominators

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

Step2: Simplify numerator of the complex fraction

Find common denominator $(x-1)^2(x-2)$:
$$\frac{(x+3)(x-2) + x(x-1)}{(x-1)^2(x-2)}$$
Expand numerator:
$$(x^2+x-6)+(x^2-x)=2x^2-6$$
Simplify numerator: $2x^2-6=2(x^2-3)$
So numerator becomes: $\frac{2(x^2-3)}{(x-1)^2(x-2)}$

Step3: Simplify denominator of the complex fraction

Find common denominator $(x-2)^2(x+2)$:
$$\frac{x(x+2) - 2(x-2)}{(x-2)^2(x+2)}$$
Expand numerator:
$$(x^2+2x)-(2x-4)=x^2+4$$
So denominator becomes: $\frac{x^2+4}{(x-2)^2(x+2)}$

Step4: Rewrite complex fraction as multiplication

Divide by a fraction = multiply by its reciprocal:
$$\frac{2(x^2-3)}{(x-1)^2(x-2)} \times \frac{(x-2)^2(x+2)}{x^2+4}$$

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: $(x-2)(x+2)=x^2-4$, so substitute:
$$\frac{2(x^2-3)(x^2-4)}{(x-1)^2(x^2+4)}$$

Answer:

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