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bakunina99 @bakunina99

August 2022 1 13 Report

ПОЖАЛУЙСТААА ПОМОГИТЕ

а) [tex]\sqrt{36} + 10\sqrt{11} + \sqrt{47 - 12 \sqrt{11}}[/tex]

б) [tex]\sqrt{87} - 16\sqrt{23} - \sqrt{39 - 8 \sqrt{23}}[/tex]

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dtnth

Verified answer

$\sqrt{36+10\sqrt{11}}+\sqrt{47-12\sqrt{11}}=\\ \sqrt{25+10\sqrt{11}+11}+\sqrt{36-12\sqrt{11}+11}=\\ \sqrt{5^2+2*5\sqrt{11}+(\sqrt{11})^2}+\sqrt{6^2-2*6\sqrt{11}+(\sqrt{11})^2}=\\ \sqrt{(5+\sqrt{11})^2}+\sqrt{(6-\sqrt{11})^2}=\\ |5+\sqrt{11}|+|6-\sqrt{11}|=\\ 5+\sqrt{11}+6-\sqrt{11}=5+6=11$

$\sqrt{87-16\sqrt{23}}-\sqrt{39-8\sqrt{23}}=\\ \sqrt{64-16\sqrt{23}+23}-\sqrt{16-8\sqrt{23}+23}=\\ \sqrt{8^2-2*8\sqrt{23}+(\sqrt{23})^2}-\sqrt{4^2-2*4\sqrt{23}+(\sqrt{23})^2}=\\ \sqrt{(8-\sqrt{23})^2}-\sqrt{(4-\sqrt{23})^2}=\\ |8-\sqrt{23}|-|4-\sqrt{23}|=\\ (8-\sqrt{23})-(\sqrt{23}-4)=8-\sqrt{23})^2-\sqrt{23}+4=8+4=12$

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