$${\text{If }}\frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 {\text{,}}$$       then the value of a and b are respectively?

$$\eqalign{ & \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {16 \times 3} + \sqrt {9 \times 2} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \frac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{\left( {4\sqrt 3 + 5\sqrt 2 } \right)\left( {4\sqrt 3 - 3\sqrt 2 } \right)}}{{48 - 18}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 + 18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 }}{{30}} + \frac{{18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{4}{{15}}\sqrt 6 + \frac{3}{5} = a + b\sqrt 6 \cr & \Rightarrow \frac{3}{5} + \frac{4}{{15}}\sqrt 6 = a + b\sqrt 6 \cr} $$
By comparing coefficients of rational and irrational parts.
$$\eqalign{ & \Rightarrow a = \frac{3}{5}{\text{ , }}b = \frac{4}{{15}} \cr & \therefore \left( {\frac{3}{5},\frac{4}{{15}}} \right) \cr} $$