ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 876

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\[\textbf{а)}\ \sqrt{2} + 5\ и\ 2 + \sqrt{5}\]

\[\sqrt{1} + 5 < \sqrt{2} + 5 < \sqrt{4} + 5\]

\[6 < \sqrt{2} + 5 < 7.\]

\[2 + \sqrt{4} < 2 + \sqrt{5} < 2 + \sqrt{9}\]

\[4 < 2 + \sqrt{5} < 5.\]

\[\sqrt{2} + 5 > 2 + \sqrt{5}.\]

\[\textbf{б)}\ \sqrt{3} - 4\ \ и\ 1 - \sqrt{5}\]

\[\sqrt{1} - 4 < \sqrt{3} - 4 < \sqrt{4} - 4\]

\[- 3 < \sqrt{3} - 4 < - 2.\]

\[1 - \sqrt{4} < 1 - \sqrt{5} < 1 - \sqrt{9}\]

\[- 1 > 1 - \sqrt{5} > - 2\]

\[- 2 < 1 - \sqrt{5} < - 1.\]

\[\sqrt{3} - 4 < 1 - \sqrt{5}\]

\[\textbf{в)}\frac{2\sqrt{3} + 23}{3}\ и\ 9\]

\[\sqrt{1} < \sqrt{3} < \sqrt{4}\]

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

\[2 \cdot 1 + 23 < 2\sqrt{3} +\]

\[+ 23 < 2 \cdot 2 + 23\]

\[25 < 2\sqrt{3} + 23 < 27\]

\[\frac{25}{3} < \frac{2\sqrt{3} + 23}{3} < 9.\]

\[\frac{2\sqrt{3} + 23}{3} < 9.\]

\[\textbf{г)}\frac{1 - \sqrt{15}}{12}\ \ и\ \ \left( - \frac{7}{8} \right)\]

\[- \sqrt{16} < - \sqrt{15} < - \sqrt{9}\]

\[- 4 < - \sqrt{15} < - 3.\]

\[\frac{1 - 4}{12} < \frac{1 - \sqrt{15}}{12} < \frac{1 - 3}{12}\]

\[- \frac{3}{12} > \frac{1 - \sqrt{15}}{12} > - \frac{2}{12}\]

\[- \frac{1}{4} > \frac{1 - \sqrt{15}}{12} > - \frac{1}{2}\]

\[- \frac{1}{2} = - \frac{4}{8}\]

\[- \frac{4}{8} > - \frac{7}{8}\]

\[\frac{1 - \sqrt{15}}{12} > - \frac{7}{8}.\]