Question

срочнооо помогите пожалуйста умоляю​

Answer 1

Ответ:

$1.\;\;a)\;\;6\sqrt{2}+3\sqrt{50}-4\sqrt{32}=6\sqrt{2}+15\sqrt{2}-16\sqrt{2}=5\sqrt{2}\\\\b)\;\;(\sqrt{32}-3\sqrt{2})*\sqrt{2}=\sqrt{64} -6=8-6=2\\\\c)\;\;(3-\sqrt{2})^2=9-6\sqrt{2}+2=11-6\sqrt{2} \\\\\\2.\frac{1}{3}\sqrt{18}\;\;>\;\;\frac{1}{2} \sqrt{4}\\\\\frac{1}{3}\sqrt{9*2}\;\;>\;\;\frac{1}{2}*2\\\\\frac{1}{3}*3\sqrt{2}\;\;>\;\; 1\\\\\sqrt{2}\;\;>\;\;1$

$3.\;\;a)\;\;\frac{\sqrt{2}-2}{\sqrt{3}-\sqrt{6}}=\frac{\sqrt{2}-(\sqrt{2})^2}{\sqrt{3}(1-\sqrt{2})}} =\frac{\sqrt{2}(1-\sqrt{2})}{\sqrt{3}(1-\sqrt{2})}}=\frac{\sqrt{2}}{\sqrt{3}} \\\\b)\;\;\frac{36-a}{6+\sqrt{a}}=\frac{(6+\sqrt{a})(6-\sqrt{a})}{(6+\sqrt{a})}=6-\sqrt{a}$

Answer 2

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