Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 189

\[1)\ y = \frac{2x^{2} - 3x + 1}{x + 1};\]

\[y^{'}(x) =\]

\[= \frac{(4x - 3)(x + 1) - \left( 2x^{2} - 3x + 1 \right)}{(x + 1)^{2}} =\]

\[= \frac{4x^{2} + 4x - 3x - 3 - 2x^{2} + 3x - 1}{(x + 1)^{2}} =\]

\[= \frac{2x^{2} + 4x - 4}{(x + 1)^{2}} = \frac{2(x^{2} + 2x - 2)}{(x + 1)^{2}}.\]

\[2)\ y = \frac{3x^{2} + 2x - 1}{2x + 1};\]

\[y^{'}(x) =\]

\[= \frac{(6x + 2)(2x + 1) - 2\left( 3x^{2} + 2x - 1 \right)}{(2x + 1)^{2}} =\]

\[= \frac{12x^{2} + 6x + 4x + 2 - 6x^{2} - 4x + 2}{(2x + 1)^{2}} =\]

\[= \frac{6x^{2} + 6x + 4}{(2x + 1)^{2}} = \frac{2(3x^{2} + 3x + 2)}{(2x + 1)^{2}}.\]

\[3)\ y = \frac{2x}{1 - x^{2}} + \frac{1}{x};\]

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

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

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

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

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

\[4)\ y = \frac{2 - x}{\sqrt{x}} + \frac{\sqrt{x}}{2 - x};\]

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

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

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

\[= \frac{(x + 2)\left( 5x - x^{2} - 4 \right)}{2x\sqrt{x}(2 - x)^{2}}.\]