Matemática: Unidad 4 - Ejercicio 25

Efectuar:
$\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$A)$	$7$
$B)$	$9$
$C)$	$2^{\textstyle \sqrt{3}}$
$D)$	$8^{\textstyle \sqrt{2}}$
$E)$	$10$

Solución

Resolviendo
	Multiplicando en aspa ambas fracciones:
		$\dfrac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})+(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
		$\dfrac{(\sqrt{3}+\sqrt{2})^{\textstyle 2}+(\sqrt{3}-\sqrt{2})^{\textstyle 2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
	Recordar:
		Binomio al cuadrado: $(a+b)^2 =a^2+2ab+b^2$
		Diferencia de cuadrados: $a^2-b^2=(a+b)(a-b)$
		$=\dfrac{(\sqrt{3}+\sqrt{2})^{\textstyle 2}+(\sqrt{3}-\sqrt{2})^{\textstyle 2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
		$=\dfrac{(\sqrt {3})^2 +2\sqrt {3}\sqrt {2}+(\sqrt {2})^2+(\sqrt {3})^2 -2\sqrt {3}\sqrt {2}+(\sqrt {2})^2}{(\sqrt {3})^2-(\sqrt {2})^2}$
	Eliminando: $2\sqrt {3}\sqrt {2}$, ya que tienen diferentes signos:
		$=\dfrac{(\sqrt {3})^2 +(\sqrt {2})^2+(\sqrt {3})^2 +(\sqrt {2})^2}{(\sqrt {3})^2-(\sqrt {2})^2}$
		$=\dfrac{3 +2+3 +2}{3-2}$
		$=\dfrac{10}{1}$
		$=10$
Respuesta:
	La solución es la Alternativa E