Question

1. $\frac{x + 5}{x^{2}-4}-\frac{x + 1}{x - 2}$

Explanation:

Step1: Factor the denominators

We know that \(x^{2}-4=(x + 2)(x - 2)\) by the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\).

Step2: Find a common denominator

The common denominator of \(\frac{x + 5}{x^{2}-4}\) and \(\frac{x + 1}{x - 2}\) is \((x + 2)(x - 2)\). Rewrite \(\frac{x + 1}{x - 2}\) with the common denominator: \(\frac{x + 1}{x - 2}=\frac{(x + 1)(x + 2)}{(x - 2)(x + 2)}=\frac{x^{2}+3x + 2}{(x + 2)(x - 2)}\). And \(\frac{x + 5}{x^{2}-4}=\frac{x + 5}{(x + 2)(x - 2)}\).

Step3: Subtract the fractions

\(\frac{x + 5}{(x + 2)(x - 2)}-\frac{x^{2}+3x + 2}{(x + 2)(x - 2)}=\frac{(x + 5)-(x^{2}+3x + 2)}{(x + 2)(x - 2)}\).
Expand the numerator: \((x + 5)-(x^{2}+3x + 2)=x + 5 - x^{2}-3x - 2=-x^{2}-2x + 3\).
So the result is \(\frac{-x^{2}-2x + 3}{(x + 2)(x - 2)}=\frac{-(x^{2}+2x - 3)}{(x + 2)(x - 2)}=\frac{-(x + 3)(x - 1)}{(x + 2)(x - 2)}\).

Answer:

\(\frac{-(x + 3)(x - 1)}{(x + 2)(x - 2)}\)