Question

racionalizar o denominador das frações √2+√3-√5/√2+√3+√5​

Answer 1

$\dfrac{\sqrt2+\sqrt3-\sqrt5}{\sqrt2+\sqrt3+\sqrt5}=\\\\\\\dfrac{\sqrt2+\sqrt3-\sqrt5}{\sqrt2+\sqrt3+\sqrt5}\cdot\dfrac{\sqrt2+\sqrt3-\sqrt5}{\sqrt2+\sqrt3-\sqrt5}=\\\\\\\dfrac{(\sqrt2+\sqrt3-\sqrt5)^2}{(\sqrt2+\sqrt3)^2-5}=\\\\\\\dfrac{(\sqrt2+\sqrt3-\sqrt5)^2}{2+2\sqrt6+3-5}=\\\\\\\dfrac{(\sqrt2+\sqrt3-\sqrt5)^2}{2\sqrt6}=\\\\\\\dfrac{(\sqrt2+\sqrt3-\sqrt5)^2}{2\sqrt6}\cdot\dfrac{\sqrt6}{\sqrt6}=\\\\\\\dfrac{\sqrt6(\sqrt2+\sqrt3-\sqrt5)^2}{12}=$

$\dfrac{\sqrt6(10+2\sqrt6-2\sqrt{10}-2\sqrt{15})}{12}=\\\\\\\dfrac{10\sqrt6+12-2\sqrt{60}-2\sqrt{90}}{12}=\\\\\\\dfrac{12+10\sqrt6-4\sqrt{15}-6\sqrt{10}}{12}=\\\\\\\boxed{\dfrac{6+5\sqrt6-2\sqrt{15}-3\sqrt{10}}{6}}$