ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 820

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

\[\frac{1}{x^{2} + 2x} - \frac{1}{x^{2} + 2x + 1} = \frac{1}{12}\]

\[y = x^{2} + 2x:\]

\[\frac{1}{y} - \frac{1}{y + 1} = \frac{1}{12}\]

\[12(y + 1) - 12y = y(y + 1)\]

\[12y + 12 - 12y = y^{2} + y\]

\[y^{2} + y - 12 = 0\]

\[D = 1 + 48 = 49\]

\[y_{1} = \frac{- 1 - 7}{2} = - 4;\]

\[y_{2} = \frac{- 1 + 7}{2} = 3.\]

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

\[x^{2} + 2x + 4 = 0\]

\[D = 4 - 8 = - 4 < 0\]

\[x \in \varnothing.\]

\[2)\ x^{2} + 2x = 3\]

\[x^{2} + 2x - 3 = 0\]

\[D = 4 + 12 = 16\]

\[x_{1} = \frac{- 2 - 4}{2} = - 3;\]

\[x_{2} = \frac{- 2 + 4}{2} = 1.\]

\[Ответ:\ - 3;\ 1.\]

\[2)\ \left( 2x - \frac{3}{x} \right) + \left( 4x^{2} + \frac{9}{x^{2}} \right) = 42\]

\[\left( 4x^{2} - 12 + \frac{9}{x^{2}} \right) + \left( 2x - \frac{3}{x} \right) - 30 = 0\]

\[\left( 2x - \frac{3}{x} \right)^{2} + \left( 2x - \frac{3}{x} \right) - 30 = 0\]

\[y = 2x - \frac{3}{x}:\]

\[y^{2} + y - 30 = 0\]

\[D = 1 + 120 = 121\]

\[y_{1} = \frac{- 1 - 11}{2} = - 6;\]

\[y_{2} = \frac{- 1 + 11}{2} = 5.\]

\[1)\ 2x - \frac{3}{x} = - 6\]

\[2x^{2} + 6x - 3 = 0\]

\[D = 36 + 24 = 60\]

\[x = \frac{- 6 \pm \sqrt{60}}{2 \bullet 2} = \frac{- 6 \pm 2\sqrt{15}}{4} =\]

\[= - \frac{3 \pm \sqrt{15}}{2}.\]

\[2)\ 2x - \frac{3}{x} = 5\]

\[2x^{2} - 5x - 3 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 - 7}{2 \bullet 2} = - \frac{1}{2};\]

\[x_{2} = \frac{5 + 7}{2 \bullet 2} = 3.\]

\[Ответ:\ - \frac{1}{2};\ 3;\ - \frac{3 \pm \sqrt{15}}{2}.\]