Question

rationalizati numitorul raportului $\frac{4}{\sqrt[4]{13}-\sqrt[4]{9} }$ va rog urgent

Answer 1

$a^4-b^4 = (a^2)^2-(b^2)^2 = (a^2-b^2)(a^2+b^2)\\ \\ a^4-b^4 = (a-b)(a+b)(a^2+b^2) \\ \\ \\\text{Deci, trebuie rationalizat cu }(a+b)(a^2+b^2).\\ \\ \\^{^{^{^{\displaystyle{(\sqrt[4]{13}+\sqrt[4]{9})(\sqrt[4]{13}^2+\sqrt[4]{9}^2)\,\Big)}}}}}\dfrac{4}{\sqrt[4]{13}-\sqrt[4]{9}} =\\ \\ \\= \dfrac{4(\sqrt[4]{13}+\sqrt[4]{9})(\sqrt[4]{13}^2+\sqrt[4]{9}^2)}{(\sqrt[4]{13}-\sqrt[4]{9})(\sqrt[4]{13}+\sqrt[4]{9})(\sqrt[4]{13}^2+\sqrt[4]{9}^2)} =$

$\\=\dfrac{4(\sqrt[4]{13}+\sqrt[4]{9})(\sqrt[4]{13}^2+\sqrt[4]{9}^2)}{\sqrt[4]{13}-\sqrt[4]{9}} =\\ \\ \\ = \dfrac{4(\sqrt[4]{13}+\sqrt[4]{9})(\sqrt[4]{13}^2+\sqrt[4]{9}^2)}{13-9} =\\\\\\=\dfrac{4(\sqrt[4]{13}+\sqrt[4]{9})(\sqrt[4]{13}^2+\sqrt[4]{9}^2)}{4}=\\ \\ \\= (\sqrt[4]{13}+\sqrt[4]{3^2})(\sqrt[4]{13}+\sqrt[4]{3^4})=\\ \\ \\= (\sqrt[4]{13}+\sqrt{3})(\sqrt{13}+3)$