ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1263

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\[1)\ \left\lbrack 1;\ 3\sqrt{2} + 2\sqrt{7} \right\rbrack;\text{\ \ }\]

\[\left\lbrack 3\sqrt{3} + 4;\ 15 \right\rbrack:\]

\[3\sqrt{2} + 2\sqrt{7} = \sqrt{\left( 3\sqrt{2} + 2\sqrt{7} \right)^{2}} =\]

\[= \sqrt{18 + 12\sqrt{14} + 28} =\]

\[= \sqrt{46 + \sqrt{2016}};\]

\[1936 < 2016 < 2025\]

\[44 < \sqrt{2016} < 45\]

\[90 < \sqrt{2016} + 46 < 91.\]

\[3\sqrt{3} + 4 = \sqrt{\left( 3\sqrt{3} + 4 \right)^{2}} =\]

\[= \sqrt{27 + 24\sqrt{3} + 16} =\]

\[= \sqrt{43 + \sqrt{1728}}.\]

\[1681 < 1728 < 1764\]

\[41 < \sqrt{1728} < 42\]

\[84 < \sqrt{1728} < 85.\]

\[Ответ:\ \ имеют.\]

\[2)\ \left( 0;\ \sqrt{27} + \sqrt{6} \right);\]

\[\left( \sqrt{48} - 1;\ 10 \right):\]

\[\sqrt{27} + \sqrt{6} = \sqrt{\left( \sqrt{27} + \sqrt{6} \right)^{2}} =\]

\[= \sqrt{27 + 2\sqrt{162}} = \sqrt{27 + \sqrt{648}}.\]

\[625 < 648 < 676\]

\[25 < \sqrt{648} < 26\]

\[52 < \sqrt{648} + 27 < 53.\]

\[\sqrt{48} - 1 = \sqrt{\left( \sqrt{48} - 1 \right)^{2}} =\]

\[= \sqrt{48 - 2\sqrt{48} + 1} =\]

\[= \sqrt{49 - \sqrt{192}}.\]

\[169 < 192 < 196\]

\[13 < \sqrt{192} < 14\]

\[- 14 < - \sqrt{192} < - 13\]

\[35 < 49 - \sqrt{192} < 36.\]

\[Ответ:\ \ имеют.\]

\[3)\ \left\lbrack 2;\ 2\sqrt{5} + 2\sqrt{6} \right\rbrack;\]

\[\left( 3\sqrt{2} + \sqrt{22};\ 11 \right):\]

\[2\sqrt{5} + 2\sqrt{6} = \sqrt{\left( 2\sqrt{5} + 2\sqrt{6} \right)^{2}} =\]

\[= \sqrt{20 + 8\sqrt{30} + 24} =\]

\[= \sqrt{44 + \sqrt{1920}}.\]

\[1849 < 1920 < 1936\]

\[43 < \sqrt{1920} < 44\]

\[87 < \sqrt{1920} + 44 < 88.\]

\[3\sqrt{2} + \sqrt{22} = \sqrt{\left( 3\sqrt{2} + \sqrt{22} \right)^{2}} =\]

\[= \sqrt{18 + 6\sqrt{44} + 22} =\]

\[= \sqrt{40 + \sqrt{1584}}.\]

\[1521 < 1584 < 1600\]

\[39 < \sqrt{1584} < 40\]

\[79 < \sqrt{1584} + 40 < 80.\]

\[Ответ:\ \ имеют.\]

\[4)\ \left\lbrack 2;\ 1 + \sqrt{3} \right\rbrack;\text{\ \ }\]

\[\left( \frac{2}{\sqrt{3} - 1};\ 11 \right):\]

\[\frac{2}{\sqrt{3} - 1} = \frac{2\left( \sqrt{3} + 1 \right)}{\left( \sqrt{3} - 1 \right)\left( \sqrt{3} + 1 \right)} =\]

\[= \frac{2\left( \sqrt{3} + 1 \right)}{3 - 1} = \sqrt{3} + 1.\]

\[Ответ:\ \ не\ имеют.\]