Question

Найти значение выражения   $frac{( sqrt{40}-7)* sqrt{7+2 sqrt{10} } }{ sqrt{15} - sqrt{6} }$

Answer 1

$sqrt{7+2sqrt{10}}=sqrt{5+2sqrt{2*5}+2}=\\sqrt{(sqrt{2})^2+2*sqrt{2}*sqrt{5}+(sqrt{5})^2}=\\sqrt{(sqrt{2}+sqrt{5})^2}=|sqrt{2}+sqrt{5}|=sqrt{5}+sqrt{2}$
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$sqrt{40}-7=-(7-sqrt{40})=-(5-sqrt{4*2*5}+2)=\\-((sqrt{5})^2-2sqrt{5}*sqrt{2}+(sqrt{2})^2)=\\-((sqrt{5}-sqrt{2})^2)$
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$sqrt{15}-sqrt{6}=sqrt{3*5}-sqrt{3*2}=\\sqrt{3}*sqrt{5}-sqrt{3}*sqrt{2}=\\sqrt{3}*(sqrt{5}-sqrt{2})$
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$frac{(sqrt{40}-7)sqrt{7+2sqrt{10}}}{sqrt{15}-sqrt{6}}=\\frac{-(sqrt{5}-sqrt{2})^2(sqrt{5}+sqrt{2})}{sqrt{3}(sqrt{5}-sqrt{2})}=\\frac{-(sqrt{5}-sqrt{2})(sqrt{5}+sqrt{2})}{sqrt{3}}=\\-frac{(sqrt{5})^2-(sqrt{2})^2}{sqrt{3}}=-frac{5-2}{sqrt{3}}=-frac{3}{sqrt{3}}=-sqrt{3}$