Question

simplify the following rational expressions:
\\(\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 denominators

First fraction denominators: $x^2-2x+1=(x-1)^2$, $x^2-3x+2=(x-1)(x-2)$; Second fraction denominators: $x^2-4x+4=(x-2)^2$, $x^2-4=(x-2)(x+2)$

Step2: Find common denominators

First sum: $\text{LCD}=(x-1)^2(x-2)$; Second difference: $\text{LCD}=(x-2)^2(x+2)$

Step3: Rewrite fractions with LCD

First sum: $\frac{(x+3)(x-2)}{(x-1)^2(x-2)} + \frac{x(x-1)}{(x-1)^2(x-2)}$; Second difference: $\frac{x(x+2)}{(x-2)^2(x+2)} - \frac{2(x-2)}{(x-2)^2(x+2)}$

Step4: Combine numerators

First sum: $\frac{(x^2+x-6)+(x^2-x)}{(x-1)^2(x-2)}=\frac{2x^2-6}{(x-1)^2(x-2)}$; Second difference: $\frac{(x^2+2x)-(2x-4)}{(x-2)^2(x+2)}=\frac{x^2+4}{(x-2)^2(x+2)}$ (Note: Corrected arithmetic in second numerator: $x(x+2)=x^2+2x$, $2(x-2)=2x-4$, so $x^2+2x-2x+4=x^2+4$)

Answer:

$\frac{(x+3)(x-2) + x(x-1)}{(x-1)^2(x-2)}$ and $\frac{x(x+2) - 2(x-2)}{(x-2)^2(x+2)}$ (simplified further as $\frac{2x^2 - 2x - 6}{(x-1)^2(x-2)}$ and $\frac{x^2 - 2}{(x-2)^2(x+2)}$)