.помогите срочно дам 30 балов. Created by mrvn0302. algebra-ru

помогите срочно дам 30 балов...

mrvn0302 @mrvn0302

July 2022 1 2 Report

.помогите срочно дам 30 балов

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NNNLLL54

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Ответ:

$1)\ \ \dfrac{1}{\sqrt8-2}=\dfrac{\sqrt8+2}{(\sqrt8-2)(\sqrt8+2)}=\dfrac{\sqrt8+2}{8-4}=\dfrac{\sqrt8+2}{4}\\\\\\2)\ \ \dfrac{9}{5+\sqrt7}=\dfrac{5-\sqrt7}{(5+\sqrt7)(5-\sqrt7)}=\dfrac{5-\sqrt7}{25-7}=\dfrac{5-\sqrt7}{18}\\\\\\3)\ \ \dfrac{4}{\sqrt5-\sqrt8}=\dfrac{4(\sqrt5+\sqrt8)}{(\sqrt5-\sqrt8)(\sqrt5+\sqrt8)}=\dfrac{4\, (\sqrt5+\sqrt8)}{5-8}=\dfrac{4\, (\sqrt5+\sqrt8)}{-3}=\\\\\\=-\dfrac{4(\sqrt5+\sqrt8)}{3}$

$4)\ \ \dfrac{18}{2\sqrt6+\sqrt{11}}=\dfrac{18\, (2\sqrt6-\sqrt{11})}{(2\sqrt6+\sqrt{11})(2\sqrt6-\sqrt{11})}=\\\\\\=\dfrac{18\, (2\sqrt6-\sqrt{11})}{4\cdot 6-11}=\dfrac{18\, (2\sqrt6-\sqrt{11})}{13}$

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nazmiranaurzalieva Вы лучший!!))

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