\[\boxed{\mathbf{587\ (587).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\left\{ \begin{matrix} y = 1\ \ \\ y = \sqrt{x} \\ \end{matrix} \right.\ \]
\[1 = \sqrt{x}\]
\[x = 1\]
\[\text{A\ }(1;1).\]
\[2)\ \left\{ \begin{matrix} y = 0,8 \\ y = \sqrt{x} \\ \end{matrix} \right.\ \]
\[0,8 = \sqrt{x}\]
\[x = {0,8}^{2}\]
\[x = 0,64\]
\[\text{B\ }(0,64;0,8).\]
\[3)\ y = - 6\]
\[y = \sqrt{x}\]
\[- 6 = \sqrt{x}\]
\[нет\ точки\ пересечения.\]
\[4)\ \left\{ \begin{matrix} y = 500 \\ y = \sqrt{x} \\ \end{matrix} \right.\ \]
\[500 = \sqrt{x}\]
\[x = 250\ 000\]
\[\text{C\ }(250000;500).\]
\[\boxed{\mathbf{5}\mathbf{8}\mathbf{7}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{a}{\sqrt{a} - 2} - \frac{4\sqrt{a} - 4}{\sqrt{a} - 2} =\]
\[= \frac{a - 4\sqrt{a} + 4}{\sqrt{a} - 2} = \frac{\left( \sqrt{a} - 2 \right)^{2}}{\sqrt{a} - 2} =\]
\[= \sqrt{a} - 2\]
\[2)\ \frac{\sqrt{m} + 1}{\sqrt{m} - 2} - \frac{\sqrt{m} + 3}{\sqrt{m}} =\]
\[= \frac{m + \sqrt{m} - m - \sqrt{m} + 6}{\sqrt{m} \cdot \left( \sqrt{m} - 2 \right)} =\]
\[= \frac{6}{m - 2\sqrt{m}}\]
\[4)\ \frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{a - 16} =\]
\[= \frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{\left( \sqrt{a} + 4 \right)\left( \sqrt{a} - 4 \right)} =\]
\[= \frac{a - 4\sqrt{a} - a}{\left( \sqrt{a} + 4 \right)\left( \sqrt{a} - 4 \right)} =\]
\[= \frac{- 4\sqrt{a}}{a - 16} = \frac{4\sqrt{a}}{16 - a}\]
\[6)\ \frac{a + \sqrt{a}}{\sqrt{b}} \cdot \frac{b}{2\sqrt{a} + 2} =\]
\[= \frac{\sqrt{a} \cdot \left( \sqrt{a} + 1 \right) \cdot b}{\sqrt{b} \cdot 2 \cdot (\sqrt{a} + 1)} = \frac{\sqrt{\text{ab}}}{2}\]