What is the best way of resolving an expression with square roots in denominator? I was resolving the following question from my textbook: 

Write each of the following expressions as a single fraction, simplifying your answer where possible:
$4 - \frac{1}{\sqrt{12}} + \frac{10}{\sqrt{3}}$

I have resolved it this way:

$ \frac{4\sqrt{12}\sqrt{3}}{\sqrt{12}\sqrt{3}} -
 \frac{\sqrt{3}}{\sqrt{12}\sqrt{3}} +
 \frac{10\sqrt{12}}{\sqrt{12}\sqrt{3}} = \frac{4\sqrt{36} -
 \sqrt{3}+10\sqrt{4}\sqrt{3}}{\sqrt{36}}=\frac{24 - \sqrt{3} +
 20\sqrt{3}}{6}$

while in the textbook it is resolved like this:

$4 - \frac{1}{\sqrt{4 \times 3}} + \frac{10}{\sqrt{3}} = 4 -
 \frac{1}{2\sqrt{3}} + \frac{10}{\sqrt{3}} = \frac{8\sqrt{3} - 1 +
 20}{2\sqrt{3}}= \frac{8\sqrt{3} + 19}{2\sqrt{3}}$

I am wondering did I do it correctly and if so, which way is better?
 A: Your answer of $$\frac{24 - \sqrt{3} +
 20\sqrt{3}}{6}=\frac{24 +
 19\sqrt{3}}{6}$$
matches the answer given by the book within a factor of $\sqrt 3\over \sqrt 3$:
$$\frac{24 +
 19\sqrt{3}}{6}=\frac{8\sqrt3+19}{2\sqrt 3}$$
Your method was effective in solving the problem, although putting $\sqrt{12}\sqrt 3$ as the denominator does not follow the "usual" method of coming up with a "least common multiple" for the final denominator; in this case, we have $(12,3)=3$, and therefore we would use $\sqrt {12}=2\sqrt 3$ as the final denominator and avoid a couple extra multiplications in the process.  But another intent in solving problems like this is often to "rationalize the denominator", which your method achieves quite effectively.  In the end, you'll want to first make sure that you understand the problem and the solution, and then follow the guidelines of the instructor you are working under: if you have a professor or teacher, match what their expectations are in the final answer form; if it's just a book or self-learning, try to match "common" notation and approaches that you see.
