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(\frac{left(4 + \frac{1}{x + 1} ight)}{left(16 - \frac{1}{(x + 1)^2} ig…

Question

(\frac{left(4 + \frac{1}{x + 1}
ight)}{left(16 - \frac{1}{(x + 1)^2}
ight)})

Explanation:

Step1: Simplify the numerator

First, simplify the numerator \(4 + \frac{1}{x + 1}\). Find a common denominator, which is \(x + 1\). So we have \(\frac{4(x + 1)+1}{x + 1}=\frac{4x + 4 + 1}{x + 1}=\frac{4x + 5}{x + 1}\).

Step2: Simplify the denominator

Next, simplify the denominator \(16-\frac{1}{(x + 1)^2}\). The common denominator is \((x + 1)^2\), so we get \(\frac{16(x + 1)^2-1}{(x + 1)^2}\). Notice that \(16(x + 1)^2-1\) is a difference of squares, where \(a = 4(x + 1)\) and \(b = 1\). So \(16(x + 1)^2-1=(4(x + 1)-1)(4(x + 1)+1)=(4x + 4 - 1)(4x + 4 + 1)=(4x + 3)(4x + 5)\). Thus, the denominator becomes \(\frac{(4x + 3)(4x + 5)}{(x + 1)^2}\).

Step3: Divide the simplified numerator by the simplified denominator

Now, we have the fraction \(\frac{\frac{4x + 5}{x + 1}}{\frac{(4x + 3)(4x + 5)}{(x + 1)^2}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes \(\frac{4x + 5}{x + 1}\times\frac{(x + 1)^2}{(4x + 3)(4x + 5)}\).

Step4: Cancel out common factors

Cancel out the common factors \(4x + 5\) and one \(x + 1\) from the numerator and denominator. We are left with \(\frac{x + 1}{4x + 3}\).

Answer:

\(\boxed{\dfrac{x + 1}{4x + 3}}\)