Question

alguien que me ayude a racionalizar esta operacion

Answer 1

$\dfrac{1}{1+\sqrt{2}-\sqrt{3}}=\dfrac{1}{\left(1+\sqrt{2}\right)-\sqrt{3}}\\ \\ \\ \dfrac{1}{1+\sqrt{2}-\sqrt{3}}\equiv \dfrac{\left(1+\sqrt{2}\right)+\sqrt{3}}{\left[\left(1+\sqrt{2}\right)-\sqrt{3}\right]\left[\left(1+\sqrt{2}\right)+\sqrt{3}\right]}\\ \\ \\ \dfrac{1}{1+\sqrt{2}-\sqrt{3}}\equiv \dfrac{1+\sqrt{2}+\sqrt{3}}{\left(1+\sqrt{2}\right)^2-\sqrt{3}^2}$

$\dfrac{1}{1+\sqrt{2}-\sqrt{3}}\equiv \dfrac{1+\sqrt{2}+\sqrt{3}}{\left(3+2\sqrt{2}\right)-3}\\ \\ \\ \dfrac{1}{1+\sqrt{2}-\sqrt{3}}\equiv \dfrac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}\\ \\ \\ \dfrac{1}{1+\sqrt{2}-\sqrt{3}}\equiv \dfrac{\left(1+\sqrt{2}+\sqrt{3}\right)\cdot \sqrt{2}}{2\sqrt{2}\cdot \sqrt{2}}\\ \\ \\ \boxed{\dfrac{1}{1+\sqrt{2}-\sqrt{3}}\equiv \dfrac{\sqrt{2}+2+\sqrt{6}}{4}}$