$a) \dfrac{x}{y}.\sqrt{\dfrac{y^4}{x^2}}+y$ với $x<0 ; y \neq 0$
$=\dfrac{x}{y}.\dfrac{y^2}{|x|}+y$
$=\dfrac{x}{y}.\dfrac{y^2}{x}+y$ ( Vì $x>0$ )
$=y+y$
$=2y$
$b) \dfrac{6y}{x^2}.\sqrt{\dfrac{x^4}{y^8}}$ với $x , y \neq 0$
$=\dfrac{6y}{x^2}.\sqrt{\dfrac{(x^2)^2}{(y^4)^2}}$
$=\dfrac{6y}{x^2}.\dfrac{x^2}{y^4}$
$=\dfrac{6x^2y}{x^2y^4}$
$=\dfrac{6}{y^3}$