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.
\[

$$\begin{align*} 1 - \frac{4}{x + 8}&=\frac{x + 8}{x + 8}-\frac{4}{x + 8}\\ &=\frac{x + 8 - 4}{x + 8}\\ &=\frac{x + 4}{x + 8} \end{align*}$$

\]

Step2: Simplify the denominator

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

$$\begin{align*} x+\frac{16}{x + 8}&=\frac{x(x + 8)}{x + 8}+\frac{16}{x + 8}\\ &=\frac{x(x + 8)+16}{x + 8}\\ &=\frac{x^{2}+8x + 16}{x + 8} \end{align*}$$

\]
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:
\[

$$\begin{align*} \frac{\frac{x + 4}{x + 8}}{\frac{(x + 4)^{2}}{x + 8}}&=\frac{x + 4}{x + 8}\times\frac{x + 8}{(x + 4)^{2}}\\ &=\frac{(x + 4)(x + 8)}{(x + 8)(x + 4)^{2}} \end{align*}$$

\]
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:

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