Question

which expression is equivalent to the following complex fraction? \\(\frac{\frac{\frac{x}{x - 3}}{x^2}}{x^2 - 9}\\)

Explanation:

Step1: Recall the rule for dividing fractions

To divide by a fraction, we multiply by its reciprocal. So, the complex fraction \(\frac{\frac{x}{x - 3}}{\frac{x^{2}}{x^{2}-9}}\) can be rewritten as \(\frac{x}{x - 3}\times\frac{x^{2}-9}{x^{2}}\).

Step2: Factor the difference of squares

We know that \(a^{2}-b^{2}=(a + b)(a - b)\). So, \(x^{2}-9=x^{2}-3^{2}=(x + 3)(x - 3)\). Substituting this into our expression, we get \(\frac{x}{x - 3}\times\frac{(x + 3)(x - 3)}{x^{2}}\).

Step3: Cancel out common factors

We can cancel out the common factors of \(x\) (one from the numerator and one from the denominator of \(x^{2}\)) and the common factor of \((x - 3)\) from the numerator and denominator. After canceling, we have \(\frac{x+3}{x}\) (we canceled one \(x\) from \(x\) and \(x^{2}\) leaving \(x\) in the denominator, and canceled \((x - 3)\) from numerator and denominator).

Answer:

\(\frac{x + 3}{x}\)