Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 749

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

\[= 1 - \frac{a - x}{a + x} = \frac{a + x - (a - x)}{a + x} =\]

\[= \frac{2x}{a + x};\]

\[a = 5;\ x = 4:\]

\[\frac{2x}{a + x} = \frac{2 \bullet 4}{5 + 4} = \frac{8}{9}.\]

\[Ответ:\ \ \frac{8}{9}.\]

\[2)\ \frac{a + \sqrt{a^{2} - x^{2}}}{a - \sqrt{a^{2} - x^{2}}} - \frac{a - \sqrt{a^{2} - x^{2}}}{a + \sqrt{a^{2} - x^{2}}} =\]

\[= \frac{4a \bullet \sqrt{a^{2} - x^{2}}}{x^{2}};\]

\[a = 3;\ x = \sqrt{5}:\]

\[\frac{4a \bullet \sqrt{a^{2} - x^{2}}}{x^{2}} =\]

\[= \frac{4 \bullet 3 \bullet \sqrt{3^{2} - \left( \sqrt{5} \right)^{2}}}{\left( \sqrt{5} \right)^{2}} =\]

\[= \frac{12\sqrt{9 - 5}}{5} = \frac{12\sqrt{4}}{5} = \frac{24}{5} = 4,8.\]

\[Ответ:\ \ 4,8.\]