Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
 A: You multiply at the numerator and at the denominator by $x^2$:
$$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}=\frac{x^2(2x+1)}{x^2 \left ( \frac{3}{x^2}+\frac{2x+1}{x}\right)}=\frac{x^2(2x+1)}{3+x(2x+1)}=\frac{x^2(2x+1)}{2x^2+x+3}$$
A: First let's start with your denominator, $$\frac{3}{x^2}+\frac{2x+1}{x}$$
We can turn this into one term by multiplying the,
 $$\frac{2x+1}{x} by \frac{x}{x}$$
and then combine terms to get, 
$$\frac{2x^2+x+3}{x^2}$$
Now if we put this back into the first expression we now have,
$$\frac{2x+1}{\frac{2x^2+x+3}{x^2}}$$
From here we simply need to multiply the numerator and denominator by x^2 and we get,
$$\frac{x^2(2x+1)}{2x^2+x+3}$$
Hope this helps!
A: The answer could be considered complete, but, as I said in my comments above, you could simplify a bit more since the numerator has a higher degree than the denominator, i.e.:$$
\frac{x^2(2x+1)}{2x^2+x+3}=\frac{2x^3+x^2}{2x^2+x+3}=\frac{2x^3+x^2+3x-3x}{2x^2+x+3}
$$$$=\frac{x(2x^2+x+3)-3x}{2x^2+x+3}=x-\frac{3x}{2x^2+x+3}$$
A: First recall that
\[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \]
And 
\[ \frac{a}{\left(\frac{c}{d}\right)} =\frac{ad}{c} \]
Then
\[
\frac{2x+1}{\frac{3}{x^{2}}+\frac{2x+1}{x}}= \frac{2x+1}{\frac{3x+x^{2}(2x+1)}{x^{3}}}= \frac{x^{3}(2x+1)}{3x+x^{2}(2x+1)}= \frac{x^{3}(2x+1)}{2x^{3}+x^{2}+3x}= \frac{x^{3}(2x+1)}{x(2x^{2}+x+3)}= \frac{x^{2}(2x+1)}{2x^{2}+x+3}
\]
