Răspuns:
Explicație pas cu pas:
[tex]... =~\dfrac{1}{\sqrt{5}+\sqrt{4}}+\dfrac{1}{\sqrt{9}+\sqrt{5}}=\dfrac{\sqrt{5}-\sqrt{4}}{(\sqrt{5}+\sqrt{4})(\sqrt{5}-\sqrt{4})}+\dfrac{\sqrt{9}-\sqrt{5}}{(\sqrt{9}+\sqrt{5})(\sqrt{9}-\sqrt{5})}}=\\=\dfrac{\sqrt{5}-\sqrt{4}}{(\sqrt{5})^2-(\sqrt{4})^2}+\dfrac{\sqrt{9}-\sqrt{5}}{(\sqrt{9})^2-(\sqrt{5})^2}=\dfrac{\sqrt{5}-\sqrt{4}}{5-4} +\dfrac{\sqrt{9}-\sqrt{5}}{9-5}=\\= \dfrac{\sqrt{5}-\sqrt{4}}{1} +\dfrac{\sqrt{9}-\sqrt{5}}{4}= \dfrac{4(\sqrt{5}-\sqrt{4})}{4} +\dfrac{\sqrt{9}-\sqrt{5}}{4}=[/tex][tex]= \dfrac{4\sqrt{5}-4*2}{4} +\dfrac{3-\sqrt{5}}{4}=\dfrac{4\sqrt{5}-4*2+3-\sqrt{5}}{4}=\dfrac{3\sqrt{5}-5}{4}.[/tex]
