Question

simplify.
\\(\dfrac{1 - \dfrac{4}{x + 8}}{x + \dfrac{16}{x + 8}}\\)

Explanation:

Step1: Simplify the numerator

First, simplify the numerator \(1 - \frac{4}{x + 8}\). We can rewrite \(1\) as \(\frac{x + 8}{x + 8}\) to have a common denominator.

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$$

Step2: Simplify the denominator

Next, simplify the denominator \(x+\frac{16}{x + 8}\). Again, rewrite \(x\) with a common denominator \((x + 8)\):

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$$

Notice that \(x^{2}+8x + 16=(x + 4)^{2}\) (by the perfect square formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a = x\), \(b = 4\)). So the denominator becomes \(\frac{(x + 4)^{2}}{x + 8}\).

Step3: Divide the simplified numerator by the simplified denominator

Now we have the complex fraction \(\frac{\frac{x + 4}{x + 8}}{\frac{(x + 4)^{2}}{x + 8}}\). Dividing by a fraction is the same as multiplying by its reciprocal:

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Then we can cancel out the common factors \((x + 4)\) and \((x + 8)\) (assuming \(x
eq - 8\) and \(x
eq - 4\) to avoid division by zero):

$$ \frac{(x + 4)(x + 8)}{(x + 8)(x + 4)^{2}}=\frac{1}{x + 4} $$

Answer:

\(\frac{1}{x + 4}\)