Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 506

\[\boxed{\mathbf{506}.}\]

\[x = \frac{2mn}{n^{2} + 1};\ \ \ m > 0;\ \ \ \]

\[0 < n < 1.\]

\[m + x = m + \frac{2mn}{n^{2} + 1} =\]

\[= \frac{mn^{2} + m + 2mn}{n^{2} + 1} =\]

\[= \frac{m\left( n^{2} + 2n + 1 \right)}{n^{2} + 1} = \frac{m(n + 1)^{2}}{n^{2} + 1}\]

\[m - x = m - \frac{2mn}{n^{2} + 1} =\]

\[= \frac{mn^{2} + m - 2mn}{n^{2} + 1} = \frac{m(n - 1)^{2}}{n^{2} + 1}\]

\[\frac{(m + x)^{\frac{1}{2}} + (m - x)^{\frac{1}{2}}}{(m + x)^{\frac{1}{2}} - (m - x)^{\frac{1}{2}}} =\]

\[= \left( \frac{\sqrt{m}(n + 1)}{\sqrt{n^{2} + 1}} + \frac{\sqrt{m}(n - 1)}{\sqrt{n^{2} + 1}} \right)\ :\]

\[:\left( \frac{\sqrt{m}(n + 1)}{\sqrt{n^{2} + 1}} - \frac{\sqrt{m}(n - 1)}{\sqrt{n^{2} + 1}} \right) =\]

\[= \frac{\sqrt{m}(n + 1 + n - 1)}{\sqrt{n^{2} + 1}} \cdot\]

\[\cdot \frac{\sqrt{n^{2} + 1}}{\sqrt{m}(n + 1 - n + 1)} = \frac{2n}{2} = n.\]