Question

simplify.
\\(\frac{\frac{x+2}{x^2}}{\frac{x^2 - 4}{x + 3}}\\), \\(x \
eq -2, 0, 2\\)
\\(\circ\\) \\(\frac{x + 3}{x^2(x - 2)}\\)
\\(\circ\\) \\(\frac{x^2 + 6}{x^4 - 4}\\)
\\(\circ\\) \\(\frac{x - 2}{x^2 + 3}\\)
\\(\circ\\) \\(\frac{3x + 5}{4x^2}\\)

Explanation:

Step1: Recall dividing fractions rule

To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction. So, \(\frac{\frac{x + 2}{x^{2}}}{\frac{x^{2}-4}{x + 3}}=\frac{x + 2}{x^{2}}\times\frac{x + 3}{x^{2}-4}\)

Step2: Factor the difference of squares

We know that \(x^{2}-4=(x + 2)(x - 2)\) (using the formula \(a^{2}-b^{2}=(a + b)(a - b)\) where \(a=x\) and \(b = 2\)). Substitute this into the expression: \(\frac{x + 2}{x^{2}}\times\frac{x + 3}{(x + 2)(x - 2)}\)

Step3: Cancel out common factors

The \((x + 2)\) terms in the numerator and denominator cancel out. We are left with \(\frac{x+3}{x^{2}(x - 2)}\)

Answer:

\(\boldsymbol{\frac{x + 3}{x^{2}(x - 2)}}\) (corresponding to the first option: A. \(\frac{x + 3}{x^{2}(x - 2)}\))