Избавьтесь от иррациональности в знаменателе 1) [tex]\frac{15}{2\sqrt{6}}[/tex] 2) [tex]\frac{1}{\sqrt{14}+\sqrt{21}+\sqrt{15}+\sqrt{10}}[/tex] 3) [tex]\frac{3}{3+\sqrt{2}-\sqrt{3}}[/tex]. Created by r091995. algebra-ru

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r091995 @r091995

October 2021 2 11 Report

Избавьтесь от иррациональности в знаменателе

[tex]\frac{15}{2\sqrt{6}}[/tex]

[tex]\frac{1}{\sqrt{14}+\sqrt{21}+\sqrt{15}+\sqrt{10}}[/tex]

[tex]\frac{3}{3+\sqrt{2}-\sqrt{3}}[/tex]

Artem112

Verified answer

$\frac{15}{2\sqrt{6}}=\frac{15\sqrt{6}}{2\sqrt{6}\sqrt{6}}=\frac{15\sqrt{6}}{12}=\frac{5\sqrt{6}}{4}$

$\frac{1}{\sqrt{14}+\sqrt{21}+\sqrt{15}+\sqrt{10}}=\frac{1}{\sqrt{2}\sqrt{7}+\sqrt{3}\sqrt{7}+\sqrt{3}\sqrt{5}+\sqrt{2}\sqrt{5}}=\frac{1}{\sqrt{7}(\sqrt{2}+\sqrt{3})+\sqrt{5}(\sqrt{2}+\sqrt{3})}=\frac{1}{(\sqrt{2}+\sqrt{3})(\sqrt{7}+\sqrt{5})}=\frac{(\sqrt{2}-\sqrt{3})(\sqrt{7}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})(\sqrt{7}+\sqrt{5})(\sqrt{2}-\sqrt{3})(\sqrt{7}-\sqrt{5})}=\frac{(\sqrt{2}-\sqrt{3})(\sqrt{7}-\sqrt{5})}{(2-3)(7-5)}=-\frac{(\sqrt{2}-\sqrt{3})(\sqrt{7}-\sqrt{5})}{2}$

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nikitaaz

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