Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1280

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

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

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

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

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

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

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

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

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

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

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

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