Question

Calculați:
$( \sqrt{4} + \sqrt{5})^{ - 1} + ( \sqrt{5} + \sqrt{9})^{ - 1}$
(radical din 4+ radical din 7)^-1+ ( radical din 7+ radical din 16)^-1=???

Answer 1

Răspuns:

Explicație pas cu pas:

$... =~\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}=$$= \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}.$