which expression is equivalent to the following complex fraction?\n$\frac{\frac{x}{x - 3}}{\frac{x^{2}}{x^{2}…

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

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

Answer

Answer:

C. $\frac{x + 3}{x}$

Explanation:

Step1: Factor the denominator

We know that $x^{2}-9=(x + 3)(x - 3)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. So the complex fraction $\frac{\frac{x}{x - 3}}{\frac{x^{2}}{x^{2}-9}}$ becomes $\frac{\frac{x}{x - 3}}{\frac{x^{2}}{(x + 3)(x - 3)}}$.

Step2: Use the rule for dividing fractions

Dividing by a fraction is the same as multiplying by its reciprocal. So $\frac{\frac{x}{x - 3}}{\frac{x^{2}}{(x + 3)(x - 3)}}=\frac{x}{x - 3}\times\frac{(x + 3)(x - 3)}{x^{2}}$.

Step3: Simplify the expression

Cancel out the common factors. The $(x - 3)$ terms cancel out, and one factor of $x$ in the numerator and denominator cancels out. We get $\frac{x+3}{x}$.