вычислить [tex](\frac{15}{\sqrt{6}+1}+ \frac{4}{\sqrt{6}-2}- \frac{12}{3-\sqrt{6}})\cdot(\sqrt{6}+11)[/tex] . Created by nikitaaz. algebra-ru

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August 2021 1 10 Report

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[tex](\frac{15}{\sqrt{6}+1}+ \frac{4}{\sqrt{6}-2}- \frac{12}{3-\sqrt{6}})\cdot(\sqrt{6}+11)[/tex]

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irkarom

$(\frac{15}{\sqrt{6}+1}+ \frac{4}{\sqrt{6}-2}- \frac{12}{3-\sqrt{6}})\cdot(\sqrt{6}+11)=\\ =(\frac{15(\sqrt{6}-2)(3-\sqrt{6})+4(\sqrt{6}+1)(3-\sqrt{6})-12(\sqrt{6}+1)(\sqrt{6}-2)}{(\sqrt{6}+1)(\sqrt{6}-2)(3-\sqrt{6})})\cdot(\sqrt{6}+11)=\\ =(\frac{15(5\sqrt{6}-12)+4(2\sqrt{6}-3)-12(4-\sqrt{6})}{18-7\sqrt6})\cdot(\sqrt{6}+11)=\\ =(\frac{75\sqrt{6}-180+8\sqrt{6}-12-48+12\sqrt{6})}{18-7\sqrt6})\cdot(\sqrt{6}+11)=\\ =(\frac{95\sqrt{6}-240}{18-7\sqrt6})\cdot(\sqrt{6}+11)=\\$

$=\frac{(95\sqrt{6}-240)(\sqrt{6}+11)}{18-7\sqrt6}=\frac{5(19\sqrt{6}-48)(\sqrt{6}+11)}{18-7\sqrt6}=\\ =\frac{5(114+209\sqrt6-48\sqrt6-528)}{18-7\sqrt6}=\frac{5(161\sqrt6-414)}{18-7\sqrt6}=\frac{5*23(7\sqrt6-18)}{18-7\sqrt6}=\\ =\frac{115(7\sqrt6-18)}{-(7\sqrt6-18)}=-115$

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