Ta có : 

a.  $\frac{y}{x}.\sqrt{\frac{x^{2}}{y^{4}}} (x>0,y\neq 0)$

<=> $\frac{y}{x}.\frac{\sqrt{x^{2}}}{\sqrt{y^{4}}}$

<=> $\frac{y}{x}.\frac{\left | x \right |}{\left | y^{2} \right |}$

Vì $x>0,y^{2}>0$ 

<=> $\frac{y}{x}.\frac{x}{y^{2}}=\frac{1}{y}$

Vậy  $\frac{y}{x}.\sqrt{\frac{x^{2}}{y^{4}}}=\frac{1}{y}$

b.   $2y^{2}.\sqrt{\frac{x^{4}}{4y^{2}}} (y<0)$

<=> $2y^{2}.\frac{\sqrt{(2x^{2})^{2}}}{\sqrt{(2y)^{2}}}$

<=> $2y^{2}.\frac{\left | x^{2} \right |}{2y}$

Vì $x^{2}\geq 0,y<0$

<=> $2y^{2}.\frac{x^{2}}{-2y}=-x^{2}y$

Vậy $2y^{2}.\sqrt{\frac{x^{4}}{4y^{2}}}=-x^{2}y$

c.  $5xy.\sqrt{\frac{25x^{2}}{y^{6}}} (x<0,y>0)$

<=>  $5xy.\frac{\sqrt{(5x)^{2}}}{\sqrt{(y^{3})^{2}}}$

<=> $5xy.\frac{\left | 5x \right |}{y^{3}}$

Vì x < 0 , y > 0 

<=> $5xy.\frac{-5x}{y^{3}}=\frac{-25x^{2}}{y^{2}}$

Vậy $5xy.\sqrt{\frac{25x^{2}}{y^{6}}} =\frac{-25x^{2}}{y^{2}}$

d.  $0,2x^{3}y^{3}.\sqrt{\frac{16}{x^{4}y^{8}}} (x\neq 0,y\neq 0)$

<=>  $0,2x^{3}y^{3}.\frac{\sqrt{16}}{\sqrt{(x^{2}y^{4})^{2}}}$

<=> $0,2x^{3}y^{3}.\frac{4}{\left | x^{2}y^{4} \right |}$

Vì $ x^{2}y^{4}>0$

<=> $0,2x^{3}y^{3}.\frac{4}{x^{2}y^{4}}=\frac{4x}{5y}$

Vậy $0,2x^{3}y^{3}.\sqrt{\frac{16}{x^{4}y^{8}}}=\frac{4x}{5y}$