\[\boxed{\text{88}\text{2}\text{\ (882).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left( \sqrt{15} + \sqrt{10} \right) \cdot 2\sqrt{5} - 5\sqrt{12} =\]
\[= \left( \sqrt{15} + \sqrt{10} \right) \cdot \sqrt{20} - \sqrt{300} =\]
\[= \sqrt{300} + \sqrt{200} - \sqrt{300} =\]
\[= \sqrt{200} = 10\sqrt{2};\]
\[\textbf{б)}\ \frac{2\sqrt{70} - 2\sqrt{28}}{3\sqrt{35} - 3\sqrt{14}} =\]
\[= \frac{2\sqrt{2} \cdot (\sqrt{35} - \sqrt{14})}{3 \cdot (\sqrt{35} - \sqrt{14})} = \frac{2\sqrt{2}}{3};\]
\[\textbf{в)}\ \left( 2\sqrt{12} - 3\sqrt{3} \right)^{2} =\]
\[= \left( 4\sqrt{3} - 3\sqrt{3} \right)^{2} = 3;\]
\[\textbf{г)}\ \frac{10 - 5\sqrt{3}}{10 + 5\sqrt{3}} + \frac{10 + 5\sqrt{3}}{10 - 5\sqrt{3}} =\]
\[= \frac{\left( 10 - 5\sqrt{3} \right)^{2} \cdot \left( 10 + 5\sqrt{3} \right)^{2}}{\left( 10 + 5\sqrt{3} \right)\left( 10 - 5\sqrt{3} \right)} =\]
\[\boxed{\text{882.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]