Question

which expression is equivalent to the following complex fraction?\
\\(\frac{\frac{x}{x - 3}}{\frac{x^3}{x^2 - 9}}\\)\
\\(\frac{x - 3}{x}\\)\
\\(\frac{x + 3}{x}\\)\
\\(\frac{x}{x + 3}\\)\
\\(\frac{x + 3}{1}\\)

Explanation:

Step1: Simplify the complex fraction

Recall that dividing by a fraction is the same as multiplying by its reciprocal. So, \(\frac{\frac{x}{x - 3}}{\frac{x^2}{x^2 - 9}}\) can be rewritten as \(\frac{x}{x - 3}\times\frac{x^2 - 9}{x^2}\).

Step2: Factor the difference of squares

Notice that \(x^2 - 9\) is a difference of squares, which factors as \((x + 3)(x - 3)\) (since \(a^2 - b^2=(a + b)(a - b)\) with \(a = x\) and \(b = 3\)). So now we have \(\frac{x}{x - 3}\times\frac{(x + 3)(x - 3)}{x^2}\).

Step3: Cancel common factors

We can cancel the \((x - 3)\) terms in the numerator and denominator, and also cancel one \(x\) from the numerator and denominator (since \(x\) in the numerator and \(x^2=x\times x\) in the denominator). After canceling, we get \(\frac{x + 3}{x}\).

Answer:

\(\boldsymbol{\frac{x + 3}{x}}\) (corresponding to the option with \(\frac{x + 3}{x}\))