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

Factor each quadratic expression:

• $x^2-2x+1=(x-1)^2$
• $x^2-3x+2=(x-1)(x-2)$
• $x^2-4x+4=(x-2)^2$
• $x^2-4=(x-2)(x+2)$

The expression becomes:
$$\frac{\frac{x+3}{(x-1)^2} + \frac{x}{(x-1)(x-2)}}{\frac{x}{(x-2)^2} - \frac{2}{(x-2)(x+2)}}$$

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 terms:
$$(x+3)(x-2)=x^2+x-6, \quad x(x-1)=x^2-x$$
Combine terms:
$$x^2+x-6+x^2-x=2x^2-6=2(x^2-3)$$
Numerator simplifies to: $\frac{2(x^2-3)}{(x-1)^2(x-2)}$

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 terms:
$$x(x+2)=x^2+2x, \quad 2(x-2)=2x-4$$
Combine terms:
$$x^2+2x-2x+4=x^2+4$$
Denominator simplifies to: $\frac{x^2+4}{(x-2)^2(x+2)}$

Step4: Divide the two simplified fractions

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

Answer:

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