Usuń niewymierność z mianownika √3+√5 / √5-√3.... Question from @muniowro

muniowro @muniowro

June 2022 2 10 Report

Usuń niewymierność z mianownika
√3+√5 / √5-√3

Obyq

Odpowiedź:

$\frac{\sqrt{3}+\sqrt{5} }{\sqrt{5}-\sqrt{3}}=\frac{3+2\sqrt{15}+5 }{5-3}=\frac{2\sqrt{15}+8 }{2} =\sqrt{15}+4$

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kontrowertyk

$\frac{\sqrt{3} +\sqrt{5} }{\sqrt{5} - \sqrt{3} } = \frac{\sqrt{3}+\sqrt{5} }{\sqrt{5}-\sqrt{3} } * \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}} = \frac{(\sqrt{3}+\sqrt{5}) }{(\sqrt{5}-\sqrt{3}) } * \frac{(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})} =$

$\frac{(\sqrt{3}+\sqrt{5})^{2}}{(\sqrt{5}-\sqrt{3}) * (\sqrt{5} + \sqrt{3}) } =\frac{(\sqrt{3}+\sqrt{5})^{2}}{5-3} = \frac{8+2\sqrt{15} }{2} = \frac{2(\sqrt{4}+\sqrt{15}) }{2} = 4 + \sqrt{15}$

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