Question

which expression is equivalent to the quotient?\\(\frac{\frac{m + 3}{m^2 - 18}}{\frac{m^2 - 1}{m + 4}}\\)\\(\frac{m - 4}{m - 3}\\)\\(\frac{1}{(m - 4)(m - 3)}\\)\\(\frac{m + 3}{m + 4}\\)\\(\frac{1}{(m + 4)(m + 3)}\\)

Explanation:

Step1: Rewrite division as multiplication

To divide by a fraction, multiply by its reciprocal. So, \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m - 4}}=\frac{m + 3}{m^2 - 16}\times\frac{m - 4}{m^2 - 9}\)

Step2: Factor the denominators

Factor \(m^2 - 16\) as a difference of squares: \(m^2 - 16=(m + 4)(m - 4)\)
Factor \(m^2 - 9\) as a difference of squares: \(m^2 - 9=(m + 3)(m - 3)\)
Substitute these into the expression: \(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m - 4}{(m + 3)(m - 3)}\)

Step3: Cancel common factors

Cancel out the common factors \(m + 3\) and \(m - 4\) from the numerator and denominator:
\(\frac{\cancel{m + 3}}{(m + 4)\cancel{(m - 4)}}\times\frac{\cancel{m - 4}}{\cancel{(m + 3)}(m - 3)}=\frac{1}{(m + 4)(m - 3)}\) Wait, no, wait. Wait, let's re - check. Wait, the original problem might have a typo? Wait, no, let's re - do the factoring. Wait, \(m^2-16=(m + 4)(m - 4)\), \(m^2 - 9=(m + 3)(m - 3)\). Then the expression is \(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m - 4}{(m + 3)(m - 3)}\). After canceling \(m + 3\) and \(m - 4\), we get \(\frac{1}{(m + 4)(m - 3)}\)? Wait, but looking at the options, there is an option \(\frac{1}{(m + 4)(m + 3)}\)? Wait, no, maybe I misread the original fraction. Wait, the original numerator of the main fraction is \(\frac{m + 3}{m^2-16}\) and the denominator is \(\frac{m^2 - 9}{m - 4}\). Wait, maybe the denominator of the second fraction is \(m + 4\) instead of \(m - 4\)? Wait, no, let's check again. Wait, if the original problem is \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m + 4}}\), then:

Rewrite as \(\frac{m + 3}{m^2 - 16}\times\frac{m + 4}{m^2 - 9}\)

Factor \(m^2-16=(m + 4)(m - 4)\), \(m^2 - 9=(m + 3)(m - 3)\)

Then \(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m + 4}{(m + 3)(m - 3)}\)

Cancel \(m + 3\) and \(m + 4\): \(\frac{1}{(m - 4)(m - 3)}\)? No, this is getting confusing. Wait, maybe the original problem has \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m + 4}}\). Let's assume that the denominator of the second fraction is \(m + 4\) (maybe a typo in the original problem's writing). Then:

\(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m + 4}{(m + 3)(m - 3)}\)

Cancel \(m + 3\) and \(m + 4\), we get \(\frac{1}{(m - 4)(m - 3)}\)? No, the options have \(\frac{1}{(m + 4)(m + 3)}\)? Wait, no, looking at the options, one of the options is \(\frac{1}{(m + 4)(m + 3)}\)? Wait, no, the user's image shows options:

First option: \(\frac{m - 4}{m - 3}\)

Second option: \(\frac{1}{(m - 4)(m - 3)}\)

Third option: \(\frac{m + 3}{m + 4}\)

Fourth option: \(\frac{1}{(m + 4)(m + 3)}\)

Wait, maybe I misread the original fraction. Let's re - express the original problem correctly. The problem is \(\frac{\frac{m + 3}{m^2-16}}{\frac{m^2 - 9}{m + 4}}\)

So, \(m^2-16=(m + 4)(m - 4)\), \(m^2 - 9=(m + 3)(m - 3)\)

Then, \(\frac{m + 3}{(m + 4)(m - 4)}\div\frac{(m + 3)(m - 3)}{m + 4}=\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m + 4}{(m + 3)(m - 3)}\)

Cancel \(m + 3\) and \(m + 4\): \(\frac{1}{(m - 4)(m - 3)}\)? No, that's not matching. Wait, maybe the original fraction is \(\frac{\frac{m + 3}{m^2-16}}{\frac{m^2 - 9}{m - 4}}\) and there is a mistake in my calculation. Wait, no, let's do it again.

\(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m - 4}}=\frac{m + 3}{m^2 - 16}\times\frac{m - 4}{m^2 - 9}=\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m - 4}{(m + 3)(m - 3)}=\frac{1}{(m + 4)(m - 3)}\)

But looking at the options, the second option is \(\frac{1}{(m - 4)(m - 3)}\) which is incorrect, and there is an option \(\frac{1}{(m + 4)(m + 3)}\) which is also incorrect. Wait, maybe the original problem has \(m^2+16\) instead of \(m^2 - 16\)? No, that's not likely. Wait, maybe I made a mistake in the sign. Wait, \(m^2-16=(m + 4)(m - 4)\), \(m^2 - 9=(m + 3)(m - 3)\). Alternatively, maybe the original fraction is \(\frac{\frac{m + 3}{m^2+16}}{\frac{m^2 - 9}{m + 4}}\), but that's not a difference of squares. Wait, perhaps the correct approach is to re - evaluate. Wait, let's look at the options again. The third option is \(\frac{m + 3}{m + 4}\). Let's see, if we have \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m - 4}}\), and we simplify:

\(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m - 4}{(m + 3)(m - 3)}=\frac{1}{(m + 4)(m - 3)}\). But if the denominator of the second fraction was \(m + 3\) instead of \(m - 3\), then we would have \(\frac{1}{(m + 4)(m + 3)}\). Alternatively, maybe the original problem is \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m + 4}}\), and after simplifying:

\(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m + 4}{(m + 3)(m - 3)}=\frac{1}{(m - 4)(m - 3)}\). But this is not matching the options. Wait, maybe the correct option is \(\frac{1}{(m + 4)(m - 3)}\) which is not listed? No, the options given are:

1. \(\frac{m - 4}{m - 3}\)

2. \(\frac{1}{(m - 4)(m - 3)}\)

3. \(\frac{m + 3}{m + 4}\)

4. \(\frac{1}{(m + 4)(m + 3)}\)

Wait, maybe I misread the original fraction. Let's assume that the original fraction is \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m + 4}}\) and there is a typo in the problem, and the denominator of the second fraction is \(m + 3\) instead of \(m - 3\). Then:

\(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m + 4}{(m + 3)(m + 3)}=\frac{1}{(m - 4)(m + 3)}\), which is not an option. Alternatively, maybe the original numerator is \(\frac{m - 3}{m^2 - 16}\). This is getting too confusing. Wait, let's start over.

The correct way to divide fractions: \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\times\frac{d}{c}\)

Given \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m - 4}}\), so \(a = m + 3\), \(b=m^2 - 16\), \(c=m^2 - 9\), \(d=m - 4\)

So, \(\frac{m + 3}{m^2 - 16}\times\frac{m - 4}{m^2 - 9}\)

Factor \(m^2 - 16=(m + 4)(m - 4)\), \(m^2 - 9=(m + 3)(m - 3)\)

Substitute: \(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m - 4}{(m + 3)(m - 3)}\)

Cancel \(m + 3\) and \(m - 4\): \(\frac{1}{(m + 4)(m - 3)}\)

But looking at the options, the second option is \(\frac{1}{(m - 4)(m - 3)}\) (incorrect), the fourth option is \(\frac{1}{(m + 4)(m + 3)}\) (incorrect), the first option \(\frac{m - 4}{m - 3}\) (incorrect), the third option \(\frac{m + 3}{m + 4}\) (incorrect). Wait, maybe there is a mistake in the problem statement. Alternatively, maybe I misread the original fraction. If the original fraction is \(\frac{\frac{m + 3}{m^2 - 16}}{\frac{m^2 - 9}{m + 4}}\), then:

\(\frac{m + 3}{(m + 4)(m - 4)}\times\frac{m + 4}{(m + 3)(m - 3)}=\frac{1}{(m - 4)(m - 3)}\) (second option). So maybe the denominator of the second fraction is \(m + 4\) instead of \(m - 4\) in the original problem. So assuming that, the correct option is the second one: \(\frac{1}{(m - 4)(m - 3)}\)

Answer:

\(\frac{1}{(m - 4)(m - 3)}\) (the second option)