Answer:
Option B is the correct answer
Step-by-step explanation:
The exact expression is:
[tex]\frac{\frac{1}{4n}-\frac{1}{2}}{\frac{n}{6}-\frac{1}{24n} }[/tex]
Taking LCM in numerator and denominator can simplify the expression. Further simplifying the expression can give us the final result, as shown below:
[tex]\frac{\frac{1-2n}{4n} }{\frac{4n^{2}-1 }{24n}} \\\\=\frac{1-2n}{4n}\times\frac{24n}{4n^{2}-1 }\\\\=\frac{1-2n}{4n}\times\frac{24n}{(2n+1)(2n-1)}\\\\ =-\frac{2n-1}{4n}\times\frac{24n}{(2n+1)(2n-1)}\\\\ =-\frac{6}{2n+1}[/tex]
Answer:
Option B. is the right answer.
Step-by-step explanation:
We have to simplify the given fraction [tex]\frac{\frac{1}{4n}-\frac{1}{2}}{\frac{n}{6}-\frac{1}{24n}}[/tex]
Now we will solve numerator first
[tex]{\frac{1}{4n}-\frac{1}{2}}[/tex] [tex]=\frac{(1-2n)}{4n}[/tex]
Then we will solve denominator
[tex]=\frac{n}{6}-\frac{1}{24n}[/tex][tex]=\frac{4n^{2}-1}{24n}[/tex]
[tex]=-\frac{-4n^{2}+1}{24n}[/tex][tex]=-\frac{(1-2n)(1+2n)}{24n}[/tex]
Now we put numerator and denominator in the form of a fraction.
[tex]-\frac{(1-2n)}{4n}\div \frac{(1-2n)(1+2n)}{24n}[/tex]
[tex]=-\frac{(1-2n)}{4n}\times \frac{24n}{(1-2n)(1+2n)}[/tex]
[tex]=-\frac{6}{2n+1}[/tex]
Therefore option B is the right answer.
