what is the common denominator of $\frac{5}{x^{2}-4}-\frac{2}{x + 2}$ in the complex fraction…

what is the common denominator of $\frac{5}{x^{2}-4}-\frac{2}{x + 2}$ in the complex fraction $\frac{\frac{2}{x - 2}-\frac{3}{x^{2}-4}}{\frac{5}{x^{2}-4}-\frac{2}{x + 2}}$?\n$(x + 2)(x - 2)$\n$x - 2$\n$(x + 2)^{2}(x - 2)$\n$x + 2$

what is the common denominator of $\frac{5}{x^{2}-4}-\frac{2}{x + 2}$ in the complex fraction $\frac{\frac{2}{x - 2}-\frac{3}{x^{2}-4}}{\frac{5}{x^{2}-4}-\frac{2}{x + 2}}$?\n$(x + 2)(x - 2)$\n$x - 2$\n$(x + 2)^{2}(x - 2)$\n$x + 2$

Answer

Answer:

A. $(x + 2)(x - 2)$

Explanation:

Step1: Factor the denominator $x^{2}-4$

Use the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. Here $a=x$ and $b = 2$, so $x^{2}-4=(x + 2)(x - 2)$.

Step2: Analyze the denominators

We have two denominators in the expression $\frac{5}{x^{2}-4}-\frac{2}{x + 2}$, which are $x^{2}-4=(x + 2)(x - 2)$ and $x + 2$.

Step3: Find the least common denominator

The least common denominator (LCD) of $(x + 2)(x - 2)$ and $x + 2$ is $(x + 2)(x - 2)$ since $(x + 2)$ is a factor of $(x + 2)(x - 2)$.