Найти значение выражения   [latex]\frac{( \sqrt{40}-7)* \sqrt{7+2 \sqrt{10} } }{ \sqrt{15} - \sqrt{6} } [/latex]

[latex]\sqrt{7+2\sqrt{10}}=\sqrt{5+2\sqrt{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}[/latex] -------------------------- [latex]\sqrt{40}-7=-(7-\sqrt{40})=-(5-\sqrt{4*2*5}+2)=\\\\-((\sqrt{5})^2-2\sqrt{5}*\sqrt{2}+(\sqrt{2})^2)=\\\\-((\sqrt{5}-\sqrt{2})^2)[/latex] -------- [latex]\sqrt{15}-\sqrt{6}=\sqrt{3*5}-\sqrt{3*2}=\\\\\sqrt{3}*\sqrt{5}-\sqrt{3}*\sqrt{2}=\\\\\sqrt{3}*(\sqrt{5}-\sqrt{2})[/latex] --------- [latex]\frac{(\sqrt{40}-7)\sqrt{7+2\sqrt{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}[/latex]