Решебник по алгебре 8 класс Мерзляк ФГОС Задание 590

\[\boxed{\mathbf{590\ (590).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 1 < \sqrt{2} < 2\]

\[2)\ 1 < \sqrt{3} < 2\]

\[3)\ 2 < \sqrt{5} < 3\]

\[4)\ 2 < \sqrt{7} < 3\]

\[5)\ 3 < \sqrt{13} < 4\]

\[6)\ 0 < \sqrt{0,98} < 1\]

\[7)\ 7 < \sqrt{59} < 8\]

\[8) - 11 < - \sqrt{115} < - 10\]

\[9) - 9 < - \sqrt{76,19} < - 8\]

\[\boxed{\mathbf{5}\mathbf{9}\mathbf{0}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}}\ :\left( \frac{\sqrt{m} + \sqrt{n}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{m} - \sqrt{n}} \right) = \frac{\sqrt{n}}{\sqrt{m}}\]

\[\frac{\sqrt{m} + {\sqrt{n}}^{\backslash\sqrt{m} - \sqrt{n}}}{\sqrt{n}} + \frac{{\sqrt{n}}^{\backslash\sqrt{n}}}{\sqrt{m} - \sqrt{n}} =\]

\[= \frac{m - n + n}{\sqrt{n}\left( \sqrt{m} - \sqrt{n} \right)} = \frac{m}{\sqrt{n}\left( \sqrt{m} - \sqrt{n} \right)}\]

\[\frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} \cdot \frac{\sqrt{n}\left( \sqrt{m} - \sqrt{n} \right)}{m} = \frac{\sqrt{n}}{\sqrt{m}}\]

\[2)\ \left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} - \frac{4\sqrt{x}}{x - 1} \right) \cdot \frac{x + \sqrt{x}}{\sqrt{x} - 1} = \sqrt{x}\]

\[\frac{\sqrt{x} + 1^{\backslash\sqrt{x} + 1}}{\sqrt{x} - 1} - \frac{4\sqrt{x}}{x - 1} =\]

\[= \frac{x + 2\sqrt{x} + 1 - 4\sqrt{x}}{\left( \sqrt{x} - 1 \right)\left( \sqrt{x} + 1 \right)} =\]

\[= \frac{x - 2\sqrt{x} + 1}{\left( \sqrt{x} - 1 \right)\left( \sqrt{x} + 1 \right)} =\]

\[= \frac{\left( \sqrt{x} - 1 \right)^{2}}{\left( \sqrt{x} - 1 \right)\left( \sqrt{x} + 1 \right)} = \frac{\sqrt{x} - 1}{\sqrt{x} + 1}\ \]

\[\frac{\sqrt{x} - 1}{\sqrt{x} + 1} \cdot \frac{x + \sqrt{x}}{\sqrt{x} - 1} =\]

\[= \frac{\left( \sqrt{x} - 1 \right)\sqrt{x}\left( \sqrt{x} + 1 \right)}{\left( \sqrt{x} - 1 \right)\left( \sqrt{x} + 1 \right)} = \sqrt{x}.\]