Answer:
[tex]\frac{\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33} }{2}[/tex]
Step-by-step explanation:
The given expression is
[tex]\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3} }[/tex]
To solve this quotient, we just have to apply a rationalization, which consists in eliminating every root in the denominator. To do so, we multiply and divide the expression by the opposite binomial of the denominator, as follows
[tex]\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3} }=\frac{\sqrt{5}-\sqrt{3} }{\sqrt{5}-\sqrt{3} }\\\\\frac{(\sqrt{6}+\sqrt{11})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}\\\\\frac{\sqrt{30}-\sqrt{18}+\sqrt{55}-\sqrt{33} }{5-3}\\ \\\frac{\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33} }{2}[/tex]
Therefore, the right answer is the second option.
