To find the product of the fractions
[tex]\[
\frac{4n}{4n-4} \cdot \frac{n-1}{n+1},
\][/tex]
we need to simplify each part of the expression before multiplying them together.
Firstly, let's simplify [tex]\(\frac{4n}{4n-4}\)[/tex]:
1. Factor out a common term from the denominator:
[tex]\[
4n - 4 = 4(n - 1).
\][/tex]
2. Now the fraction becomes:
[tex]\[
\frac{4n}{4(n - 1)} = \frac{4n}{4(n - 1)}.
\][/tex]
3. We can cancel out the common factor of 4:
[tex]\[
\frac{4n}{4(n - 1)} = \frac{n}{n - 1}.
\][/tex]
So, the simplified form of [tex]\(\frac{4n}{4n-4}\)[/tex] is [tex]\(\frac{n}{n-1}\)[/tex].
Secondly, consider the fraction [tex]\(\frac{n-1}{n+1}\)[/tex]:
The fraction [tex]\(\frac{n-1}{n+1}\)[/tex] is already in its simplest form.
Now, we multiply the two simplified fractions together:
[tex]\[
\frac{n}{n-1} \cdot \frac{n-1}{n+1}.
\][/tex]
Notice that [tex]\((n-1)\)[/tex] appears once in the numerator and once in the denominator, so they cancel each other out:
[tex]\[
\frac{n}{n-1} \cdot \frac{n-1}{n+1} = \frac{n \cancel{(n-1)}}{\cancel{(n-1)}(n+1)} = \frac{n}{n+1}.
\][/tex]
Therefore, the product of the given fractions simplifies to:
[tex]\[
\boxed{\frac{n}{n+1}}.
\][/tex]
