Respuesta :

let's use the conjugate of the denominator and multiply top and bottom by it, recall the conjugate of a binomial is simply the same binomial with a different sign in between.

[tex]\bf \cfrac{2\sqrt{x}-3\sqrt{y}}{\sqrt{x}+\sqrt{y}}\cdot \cfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}\implies \cfrac{2\sqrt{x}\sqrt{x}-2\sqrt{x}\sqrt{y}~~-~~3\sqrt{x}\sqrt{y}+3\sqrt{y}\sqrt{y}}{\underset{\textit{difference of squares}}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}} \\\\\\ \cfrac{2\sqrt{x^2}-2\sqrt{xy}-3\sqrt{xy}+3\sqrt{y^2}}{(\sqrt{x})^2-(\sqrt{y})^2}\implies \cfrac{2x-5\sqrt{xy}+3y}{x-y}[/tex]

Answer:

[tex]\dfrac{2x-5\sqrt{xy}+3y}{x-y}\\[/tex]

Step-by-step explanation:

In Rationalize the denominator we multiply both numerator and denominator by the conjugate of denominator.

In Conjugate we change the sign of middle operator.

Example: Congugate of (a + b) = a - b

Now Solving the given expression,

[tex]\dfrac{2\sqrt{x} - 3\sqrt{y}}{\sqrt{x} + \sqrt{y}} = \dfrac{2\sqrt{x} - 3\sqrt{y}}{\sqrt{x} + \sqrt{y}}\times \dfrac{\sqrt{x} - \sqrt{y}}{\sqrt{x} - \sqrt{y}}\\\\\Rightarrow \dfrac{(2\sqrt{x} - 3\sqrt{y})(\sqrt{x} - \sqrt{y})}{( \sqrt{x} + \sqrt{y}){(\sqrt{x} - \sqrt{y}})}\ \ \ \ \ \ \ \ \ \ \ [\because (a-b)(a+b)=(a^{2} +b^{2})]\\\Rightarrow \dfrac{2x-2\sqrt{xy}-3\sqrt{xy}+3y}{x-y}\\\\ \Rightarrow \dfrac{2x-5\sqrt{xy}+3y}{x-y}\\[/tex]