Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 750

\[1)\ \frac{x^{\frac{1}{2}}}{1 + x^{\frac{1}{2}}} \bullet \left( \frac{x^{\frac{1}{2}}}{1 - x^{\frac{1}{2}}} - \frac{1}{x^{\frac{1}{2}} - x} \right) =\]

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

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

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

\[= - \frac{1 - x}{1 - x} = - 1.\]

\[2)\ \frac{m + 2m^{\frac{1}{2}} + 1}{2m^{\frac{1}{2}}} \bullet \left( \frac{2m^{\frac{1}{2}}}{m^{\frac{1}{2}} - 1} - \frac{4m^{\frac{1}{2}}}{m - 1} \right) =\]

\[= \frac{m + 2\sqrt{m} + 1}{2\sqrt{m}} \bullet \left( \frac{2\sqrt{m}}{\sqrt{m} - 1} - \frac{4\sqrt{m}}{m - 1} \right) =\]

\[= \frac{\left( \sqrt{m} + 1 \right)^{2}}{2\sqrt{m}} \bullet \frac{2\sqrt{m} \bullet \left( \sqrt{m} + 1 \right) - 4\sqrt{m}}{\left( \sqrt{m} - 1 \right)\left( \sqrt{m} + 1 \right)} =\]

\[= \frac{\sqrt{m} + 1}{2\sqrt{m}} \bullet \frac{2m + 2\sqrt{m} - 4\sqrt{m}}{\sqrt{m} - 1} =\]

\[= \frac{\sqrt{m} + 1}{2\sqrt{m}} \bullet \frac{2m - 2\sqrt{m}}{\sqrt{m} - 1} =\]

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

\[= \sqrt{m} + 1.\]