как решить дам 25 баллов [tex]\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}} -\sqrt{1,2}*\sqrt{20} [/tex]. Created by sona238. algebra-ru

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$\frac{ \sqrt{5+2 \sqrt{6} } }{ \sqrt{5-2 \sqrt{6} } }- \sqrt{1,2}* \sqrt{20} = \frac{ \sqrt{5+2 \sqrt{6} } * \sqrt{5+2 \sqrt{6} } }{ \sqrt{5-2 \sqrt{6} }* \sqrt{5+2 \sqrt{6} } } - \sqrt{1,2*20} } = \frac{( \sqrt{5+2 \sqrt{6} }) ^{2} }{ \sqrt{5 ^{2} -(2 \sqrt{6} ) ^{2} } } -$ $\sqrt{24}= \frac{5+2 \sqrt{6} }{ \sqrt{25-24} }- \sqrt{4*6} = \frac{5+2 \sqrt{6} }{1}-2 \sqrt{6}=5+2 \sqrt{6} -2 \sqrt{6} = 5$

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$\cfrac{ \sqrt{5+2 \sqrt{6} } }{\sqrt{5-2 \sqrt{6} }} - \sqrt{1.2}* \sqrt{20} = \cfrac{ \sqrt{ (\sqrt{3})^2 +2 \sqrt{2} \sqrt{3}+( \sqrt{2} )^2 } }{\sqrt{(\sqrt{3})^2 -2 \sqrt{2} \sqrt{3}+( \sqrt{2} )^2 }} - \sqrt{1.2*20} = \\\\\\ =\cfrac{ \sqrt{ ( \sqrt{3}+ \sqrt{2} )^2 } }{\sqrt{( \sqrt{3}- \sqrt{2} )^2 }} - \sqrt{24} = \cfrac{\sqrt{3}+ \sqrt{2}}{\sqrt{3}- \sqrt{2}} - \sqrt{24} =$

$=\cfrac{(\sqrt{3}+ \sqrt{2})(\sqrt{3}+ \sqrt{2})}{(\sqrt{3}- \sqrt{2})(\sqrt{3}+ \sqrt{2})} - \sqrt{24} =\cfrac{3+2\sqrt{6 }+2 }{3-2} - 2\sqrt{6} = \\\\\\ =5+2 \sqrt{6}-2 \sqrt{6}=5$

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