Question

the image contains two rational expression operations:

1. addition: \\(\frac{x + 3}{x^2 - 2x + 1} + \frac{x}{x^2 - 3x + 2}\\)
2. subtraction: \\(\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 first

Find common denominator for numerator terms: $(x-1)^2(x-2)$
$$\frac{(x+3)(x-2) + x(x-1)}{(x-1)^2(x-2)}$$
Expand numerator:
$(x+3)(x-2)=x^2+x-6$, $x(x-1)=x^2-x$
Sum: $x^2+x-6+x^2-x=2x^2-6=2(x^2-3)$
Numerator simplified: $\frac{2(x^2-3)}{(x-1)^2(x-2)}$

Step3: Simplify denominator next

Find common denominator for denominator terms: $(x-2)^2(x+2)$
$$\frac{x(x+2) - 2(x-2)}{(x-2)^2(x+2)}$$
Expand numerator:
$x(x+2)=x^2+2x$, $2(x-2)=2x-4$
Difference: $x^2+2x-(2x-4)=x^2+4$
Denominator simplified: $\frac{x^2+4}{(x-2)^2(x+2)}$

Step4: Divide simplified numerator/denominator

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

Answer:

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