Answer :
Step-by-step explanation:
[tex]Domain:\ x>0\ \wedge\ y>0[/tex]
[tex]\dfrac{2\sqrt{x}-3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\qquad|\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{2\sqrt{x}-3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{(2\sqrt{x}-3\sqrt{y})(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}\\\\\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=\dfrac{(2\sqrt{x})(\sqrt{x})+(2\sqrt{x})(-\sqrt{y})+(-3\sqrt{y})(\sqrt{x})+(-3\sqrt{y})(-\sqrt{y})}{(\sqrt{x})^2-(\sqrt{y})^2}[/tex]
[tex]\text{use}\ (\sqrt{a})^2=a\ \text{and}\ (\sqrt{a})(\sqrt{a})=a\ \text{for}\ a\geq0\\\text{and}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\=\dfrac{2x-2\sqrt{xy}-3\sqrt{xy}+3y}{x-y}\qquad|\text{combine like terms}\\\\=\dfrac{2x+3y-5\sqrt{xy}}{x-y}[/tex]