Ta có :

a.   $\sqrt{\frac{a}{b}}+\sqrt{ab}+\frac{a}{b}\sqrt{\frac{b}{a}}$  với a > 0 , b > 0 

=  $\sqrt{\frac{ab}{b^{2}}}+\sqrt{ab}+\frac{a}{b}\sqrt{\frac{ab}{a^{2}}}$

=  $\frac{\sqrt{ab}}{b}+\sqrt{ab}+\frac{\sqrt{ab}}{b}=\frac{(b+2)\sqrt{ab}}{b}$

Vậy $\sqrt{\frac{a}{b}}+\sqrt{ab}+\frac{a}{b}\sqrt{\frac{b}{a}}=\frac{(b+2)\sqrt{ab}}{b}$

b.   $\sqrt{\frac{m}{1-2x+x^{2}}}.\sqrt{\frac{4m-8mx+4mx^{2}}{81}}$  với m > 0  và $x\neq 1$

=  $\sqrt{\frac{m}{(1-x)^{2}}}.\sqrt{\frac{4m(1-2x+x^{2})}{81}}$

=  $\sqrt{\frac{m}{(1-x)^{2}}}.\sqrt{\frac{4m(1-x)^{2}}{81}}$

=  $\sqrt{\frac{4m^{2}}{81}}=\frac{\sqrt{4m^{2}}}{\sqrt{81}}=\frac{\left | 2m \right |}{9}=\frac{2m}{9}$

Vậy  $\sqrt{\frac{m}{1-2x+x^{2}}}.\sqrt{\frac{4m-8mx+4mx^{2}}{81}}=\frac{2m}{9}$