Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 97

\[1)\ arctg\ 2\sqrt{3}\ и\ arctg\ 3\sqrt{2};\]

\[\frac{2\sqrt{3}}{3\sqrt{2}} = \frac{\sqrt{12}}{\sqrt{18}} < 1;\]

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

\[2)\ arctg\left( - \frac{1}{\sqrt{2}} \right)\ и\ \text{arctg}\left( - \frac{1}{\sqrt{5}} \right);\]

\[\frac{1}{\sqrt{2}}\ :\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{2}} > 1;\]

\[\text{arctg}\left( - \frac{1}{\sqrt{2}} \right) < arctg\left( - \frac{1}{\sqrt{5}} \right).\]

\[3)\ arcctg\ \sqrt{5}\ и\ arcctg\ \sqrt{7};\]

\[\frac{\sqrt{5}}{\sqrt{7}} < 1;\]

\[\text{atcctg\ }\sqrt{5} > arcctg\ \sqrt{7}.\]

\[4)\ arcctg\left( - \frac{2}{\sqrt{3}} \right)\ и\ arcctg\left( - \sqrt{2} \right);\]

\[\frac{2}{\sqrt{3}}\ :\sqrt{2} = \frac{2}{\sqrt{6}} = \frac{\sqrt{4}}{\sqrt{6}} < 1;\]

\[\text{arcctg}\left( - \frac{2}{\sqrt{3}} \right) < arcctg\left( - \sqrt{2} \right).\]