a, đk x ≠ y ; xy>0
b,
\(B=\dfrac{x\sqrt{x}-\sqrt{xy}+y\sqrt{y}+\sqrt{xy}}{x-\sqrt{xy}+y}.\dfrac{1}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{x\sqrt{x}+y\sqrt{y}}{x-\sqrt{xy}+y}.\dfrac{1}{\sqrt{x}+\sqrt{y}}=\dfrac{\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3}{x-\sqrt{xy}+y}.\dfrac{1}{\sqrt{x}+\sqrt{y}}=1\)
