Question

the image shows two rational expressions: the first is \\(\frac{x + 3}{x^2 - 2x + 1} + \frac{x}{x^2 - 3x + 2}\\) and the second is \\(\frac{x}{x^2 - 4x + 4} - \frac{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 $(x-1)^2(x-2)$:
$$\frac{(x+3)(x-2)}{(x-1)^2(x-2)} + \frac{x(x-1)}{(x-1)^2(x-2)}$$
Expand numerators:
$$(x+3)(x-2)=x^2+x-6, \quad x(x-1)=x^2-x$$
Add numerators:
$$x^2+x-6+x^2-x=2x^2-6=2(x^2-3)$$
Simplified numerator: $\frac{2(x^2-3)}{(x-1)^2(x-2)}$

Step3: Simplify denominator of the complex fraction

Find common denominator $(x-2)^2(x+2)$:
$$\frac{x(x+2)}{(x-2)^2(x+2)} - \frac{2(x-2)}{(x-2)^2(x+2)}$$
Expand numerators:
$$x(x+2)=x^2+2x, \quad 2(x-2)=2x-4$$
Subtract numerators:
$$x^2+2x-(2x-4)=x^2+4$$
Simplified denominator: $\frac{x^2+4}{(x-2)^2(x+2)}$

Step4: Divide simplified 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:
$$\frac{2(x^2-3)(x^2-4)}{(x-1)^2(x^2+4)}$$

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: $x
eq 1, 2, -2$