\[\boxed{\mathbf{571\ (571).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\frac{\sqrt{3} + \sqrt{2} + \sqrt{2} + 1}{\left( \sqrt{2} + 1 \right)\left( \sqrt{3} + \sqrt{2} \right)} =\]
\[= \frac{1 + 2\sqrt{2} + \sqrt{3}}{\left( \sqrt{2} + 1 \right)\left( \sqrt{3} + \sqrt{2} \right)}\]
\[= \frac{4\sqrt{3} + 3\sqrt{6} + 5\sqrt{2} + 7}{4\sqrt{3} + 3\sqrt{6} + 5\sqrt{2} + 7} = 1\]
\[3)1 + \frac{1}{\sqrt{5} + \sqrt{4}} = \frac{\sqrt{5} + \sqrt{4} + 1}{\sqrt{5} + \sqrt{4}} =\]
\[= - 1 + \sqrt{5}\]
\[4)\ Таким\ образом,\ последняя\ \]
\[сумма\ будет\ иметь\ результат:\]
\[- 1 + \sqrt{100} = - 1 + 10 = 9\]
\[Ответ:9.\ \]
\[\boxed{\mathbf{5}\mathbf{7}\mathbf{1}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ 15 - x^{2} =\]
\[= (\sqrt{15} - x)(\sqrt{15} + x)\]
\[2)\ 49x² - 2 =\]
\[= (7x - \sqrt{2})(7x + \sqrt{2})\]
\[3)\ 36p - 64q =\]
\[= (6\sqrt{p} - 8\sqrt{q})(6\sqrt{p} + 8\sqrt{q})\]
\[4)\ c - 100 =\]
\[= (\sqrt{c} - 10)(\sqrt{c} + 10)\]
\[5)\ a - 8b\sqrt{a} + 16b^{2} =\]
\[= (\sqrt{a} - 4b)²\]
\[6)\ m + 2\sqrt{\text{mn}} + n =\]
\[= (\sqrt{m} + \sqrt{n})²\]
\[7)\ a - 4\sqrt{a} + 4 = (\sqrt{a} - 2)²\]
\[8)\ 5 + \sqrt{5} = \sqrt{5} \cdot (\sqrt{5} + 1)\]
\[9)\sqrt{3p} - p = \sqrt{p} \cdot (\sqrt{3} - \sqrt{p})\]
\[10)\ \sqrt{12} + \sqrt{32} = 2\sqrt{3} + 4\sqrt{2} =\]
\[= 2 \cdot (\sqrt{3} + 2\sqrt{2})\ \]