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

Step2: Simplify numerator of the complex fraction

Find common denominator $(x-1)^2(x-2)$ for numerator terms:
$$\frac{x+3}{(x-1)^2} + \frac{x}{(x-1)(x-2)} = \frac{(x+3)(x-2) + x(x-1)}{(x-1)^2(x-2)}$$
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 $(x-2)^2(x+2)$ for denominator terms:
$$\frac{x}{(x-2)^2} - \frac{2}{(x-2)(x+2)} = \frac{x(x+2) - 2(x-2)}{(x-2)^2(x+2)}$$
Expand numerator:
$$(x^2+2x) - (2x-4) = x^2 + 4$$
So denominator becomes: $\frac{x^2+4}{(x-2)^2(x+2)}$

Step4: Divide numerator by denominator

Dividing by a fraction is multiplying by its reciprocal:
$$\frac{2(x^2-3)}{(x-1)^2(x-2)} \times \frac{(x-2)^2(x+2)}{x^2+4}$$
Cancel common factor $(x-2)$:
$$\frac{2(x^2-3)(x-2)(x+2)}{(x-1)^2(x^2+4)}$$
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 $\frac{2(x^4-7x^2+12)}{(x^2-2x+1)(x^2+4)}$ (where $x
eq 1, 2, -2$)