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

Step2: Simplify numerator first

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

Step3: Simplify denominator next

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

Step4: Rewrite as multiplication 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 $(x-2)$:
$\frac{2(x^2-3)(x-2)(x+2)}{(x-1)^2(x^2+4)}$
Combine $(x-2)(x+2)=x^2-4$:
$\frac{2(x^2-3)(x^2-4)}{(x-1)^2(x^2+4)}$

Answer:

$\frac{2(x^2-3)(x^2-4)}{(x-1)^2(x^2+4)}$ (or expanded form $\frac{2(x^4-7x^2+12)}{(x-1)^2(x^2+4)}$)
Note: $x
eq 1, 2, -2$ to avoid division by zero.