Simplify the given expression-
$\dfrac{{\sqrt 6 }}{{\sqrt 5 + \sqrt 3 }} + \dfrac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \dfrac{{4\sqrt 3 }}{{6 + \sqrt 2 }}$

Hint: In these type of questions where denominators are irrational with plus and minus sign in between, we proceed by rationalising the denominator by multiplying the denominator and numerator by the same term of the denominator but opposite in sign from which we get the denominator as an integer which makes our problem easy.

Complete step-by-step answer:
In this question, we are given to simplify $\dfrac{{\sqrt 6 }}{{\sqrt 5 + \sqrt 3 }} + \dfrac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \dfrac{{4\sqrt 3 }}{{6 + \sqrt 2 }}$, so we need to rationalise the denominators of all the terms and then take the $LCM$ of all the denominators to convert it into a single term.
So let us proceed:
$\dfrac{{\sqrt 6 }}{{\sqrt 5 + \sqrt 3 }} + \dfrac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \dfrac{{4\sqrt 3 }}{{6 + \sqrt 2 }}$
$\dfrac{{\sqrt 6 (\sqrt 5 - \sqrt 3 )}}{{(\sqrt 5 + \sqrt 3 )(\sqrt 5 - \sqrt 3 )}} + \dfrac{{3\sqrt 2 (\sqrt 6 - \sqrt 3 )}}{{(\sqrt 6 + \sqrt 3 )(\sqrt 6 - \sqrt 3 )}} - \dfrac{{4\sqrt 3 (6 - \sqrt 2 )}}{{(6 + \sqrt 2 )(6 - \sqrt 2 )}}$$ - - - - - - (1)$
So as we know the formula that $(a + b)(a - b) = {a^2} - {b^2}$
Let us apply it in the equation (1)
$\dfrac{{\sqrt 6 (\sqrt 5 - \sqrt 3 )}}{{(5 - 3)}} + \dfrac{{3\sqrt 2 (\sqrt 6 - \sqrt 3 )}}{{(6 - 3)}} - \dfrac{{4\sqrt 3 (6 - \sqrt 2 )}}{{(36 - 2)}}$$ - - - - - (2)$
Now we also know that $\sqrt a .\sqrt b = \sqrt {ab} $
Hence we can use this in the numerators of the equation (2)
$\dfrac{{\sqrt {30} - \sqrt {18} }}{{(5 - 3)}} + \dfrac{{3\sqrt {12} - 3\sqrt 6 }}{{(6 - 3)}} - \dfrac{{24\sqrt 3 - 4\sqrt 6 }}{{(36 - 2)}}$
$\dfrac{{\sqrt {30} - \sqrt {18} }}{2} + \dfrac{{3\sqrt {12} - 3\sqrt 6 }}{3} - \dfrac{{24\sqrt 3 - 4\sqrt 6 }}{{34}}$
$\dfrac{{\sqrt {30} - \sqrt {18} }}{2} + \dfrac{{2\sqrt 3 - \sqrt 6 }}{1} - \dfrac{{24\sqrt 3 - 4\sqrt 6 }}{{34}}$$ - - - - - (3)$
Taking $LCM$
So we know that the $LCM$ least common multiple of all the denominators I that is $2,1,34$ which is $34$
Hence making all the denominators $34$
So we need to multiply denominator and numerator by $17,34,1$in the first, second, third term of the equation (3)
$\dfrac{{17\sqrt {30} - 17\sqrt {18} }}{{34}} + \dfrac{{68\sqrt 3 - 34\sqrt 6 }}{{34}} - \dfrac{{24\sqrt 3 - 4\sqrt 6 }}{{34}}$
$\dfrac{{17\sqrt {30} - 51\sqrt 2 + 68\sqrt 3 - 34\sqrt 6 - 24\sqrt 3 + 4\sqrt 6 }}{{34}}$
$\dfrac{{17\sqrt {30} - 51\sqrt 2 + 44\sqrt 3 - 30\sqrt 6 }}{{34}}$
which is its simplified form.

Note: Here to solve these kinds of questions, we should keep in mind the concept of rationalising the denominator and of making the common denominator by taking the $LCM$ and by this way our problem will be solved easily.