Question

1. \\(\frac{x^2 + 6x + 9}{x^2 - 9} cdot \frac{x - 3}{x + 3}\\)

Explanation:

Step1: Factor numerators and denominators

First, factor the quadratic expressions. The numerator \(x^2 + 6x + 9\) is a perfect square trinomial, so it factors to \((x + 3)^2\). The denominator \(x^2 - 9\) is a difference of squares, so it factors to \((x + 3)(x - 3)\). So the expression becomes:
\(\frac{(x + 3)^2}{(x + 3)(x - 3)} \cdot \frac{x - 3}{x + 3}\)

Step2: Cancel common factors

Now, cancel out the common factors in the numerators and denominators. We can cancel \((x + 3)\) from the first fraction's numerator and denominator, and we can also cancel \((x - 3)\) from the first fraction's denominator and the second fraction's numerator. After canceling, we have:
\(\frac{(x + 3)}{(x - 3)} \cdot \frac{(x - 3)}{(x + 3)}\)
Then, multiplying the remaining terms, \((x + 3)\) and \((x - 3)\) in the numerators and denominators will also cancel out, leaving us with 1 (as long as \(x
eq\pm3\) to avoid division by zero).

Answer:

\(1\) (for \(x
eq\pm3\))