Question

\frac{9}{
\frac{1}{3} + \frac{-1}{u}
} =

Explanation:

Step1: Simplify the denominator

First, we need to simplify the denominator \(\frac{1}{3}+\frac{-1}{u}\). To add these two fractions, we find a common denominator, which is \(3u\). So we rewrite each fraction with the common denominator:
\(\frac{1}{3}=\frac{u}{3u}\) and \(\frac{-1}{u}=\frac{-3}{3u}\).
Then, \(\frac{1}{3}+\frac{-1}{u}=\frac{u}{3u}+\frac{-3}{3u}=\frac{u - 3}{3u}\).

Step2: Divide by the simplified denominator

Now, we have \(\frac{9}{\frac{u - 3}{3u}}\). Dividing by a fraction is the same as multiplying by its reciprocal. So we multiply 9 by the reciprocal of \(\frac{u - 3}{3u}\), which is \(\frac{3u}{u - 3}\).
So, \(9\times\frac{3u}{u - 3}=\frac{27u}{u - 3}\) (assuming \(u
eq3\) and \(u
eq0\) to avoid division by zero).

Answer:

\(\frac{27u}{u - 3}\) (for \(u
eq0\) and \(u
eq3\))