Rationalize $\frac{5\sqrt{ 3 }-4\sqrt{ 2 }}{4\sqrt{ 3 }+3\sqrt{ 2 }}$  and choose the option which represents the rationalized form correctly.

We are given a fraction $\frac{5\sqrt{ 3 }-4\sqrt{ 2 }}{4\sqrt{ 3 }+3\sqrt{ 2 }}$  which is to be rationalized. 
To rationalize the fraction, we multiply both the numerator and the denominator by $4\sqrt{3}-3\sqrt{2}$  and simplify using the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$ .

$\frac{5\sqrt{ 3 }-4\sqrt{ 2 }}{4\sqrt{ 3 }+3\sqrt{ 2 }}$  =  $\frac{5\sqrt{3}-4\sqrt{2}}{4\sqrt{3}+3\sqrt{2}} \times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}$ 

$\begin{array}{l}=\frac{5\sqrt{3}\left(4\sqrt{3}-3\sqrt{2}\right)-4\sqrt{2}\left(4\sqrt{3}-3\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2}\\ \\ =\frac{20\times 3-15\sqrt{6}-16\sqrt{6}+12\times 2}{48-18}\\ \\ =\frac{60+24-31\sqrt{6}}{30}\\ \\ =\frac{84-31\sqrt{6}}{30}\end{array}$ 
So, rationalizing the denominator of $\frac{5\sqrt{ 3 }-4\sqrt{ 2 }}{4\sqrt{ 3 }+3\sqrt{ 2 }}$ , the result is obtained as $\frac{84-31\sqrt{ 6 }}{30}$ .
Hence, the correct option is $\frac{84-31\sqrt{ 6 }}{30}$.