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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 391
Задание 391
\[\boxed{\mathbf{391}.}\]
\[1)\ \frac{x^{2}}{x + 2} + \frac{2x^{2}(x - 2)}{x - 3} =\]
\[= \frac{3x^{2} + 19x + 6}{x^{2} - x - 6}\text{\ \ }\]
\[\ | \cdot (x + 2)(x - 3)\]
\[x^{2} - x - 6 = (x + 2)(x - 3)\]
\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]
\[x_{1} = 3;\ \ x_{2} = - 2.\]
\[ОДЗ:\ \ x \neq - 2;\ \ x \neq 3.\]
\[x^{2}(x - 3) +\]
\[+ 2x^{2}(x - 2)(x + 2) =\]
\[= 3x^{2} + 19x + 6\]
\[x^{3} - 3x^{2} + 2x^{4} - 8x^{2} - 3x^{2} -\]
\[- 19x - 6 = 0\]
\[2x^{4} + x^{3} - 14x^{2} - 19x - 6 = 0\]
\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 6.\]
| \[2\] | \[1\] | \[- 14\] | \[- 19\] | \[- 6\] | |
|---|---|---|---|---|---|
| \[- 2\] | \[2\] | \[- 3\] | \[- 8\] | \[- 3\] | \[0\] |
| \[- 1\] | \[2\] | \[- 5\] | \[- 3\] | \[0\] |
\[P(x) = (x + 2)(x + 1)\left( 2x^{2} - 5x - 3 \right) = 0.\]
\[2x^{2} - 5x - 3 = 0\]
\[D = 25 + 24 = 49\]
\[x_{1} = \frac{5 + 7}{4} =\]
\[= 3\ (не\ подъодит\ по\ ОДЗ);\ \ \]
\[x_{2} = \frac{5 - 7}{4} = - 0,5.\]
\[Ответ:x = - 1;\ \ x = - 0,5.\]
\[2)\ \frac{2x^{3}}{x + 2} + \frac{x^{2}}{x - 1} =\]
\[= \frac{8x^{2} - 7x + 2}{x^{2} + x - 2}\text{\ \ \ \ \ }\]
\[| \cdot (x + 2)(x - 1)\]
\[x^{2} + x - 2 = (x + 2)(x - 1)\]
\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = - 2;\ \ x_{3} = 1.\]
\[ОДЗ:\ \ x = 1;\ \ x \neq - 2.\]
\[2x^{3}(x - 1) + x^{2}(x + 2) =\]
\[= 8x^{2} - 7x + 2\]
\[2x^{4} - 2x^{3} + x^{3} + 2x^{2} -\]
\[- 8x^{2} + 7x - 2 = 0\]
\[2x^{4} - x^{3} - 6x^{2} + 7x - 2 = 0\]
\[Делители:\ \pm 1;\ \pm 2.\]
| \[2\] | \[- 1\] | \[- 6\] | \[7\] | \[- 2\] | |
|---|---|---|---|---|---|
| \[- 2\] | \[2\] | \[- 5\] | \[4\] | \[- 1\] | \[0\] |
| \[1\] | \[2\] | \[- 3\] | \[1\] | \[0\] | |
| \[0,5\] | \[2\] | \[- 2\] | \[0\] |
\[P(x) = (x + 2)(x - 1)(x - 0,5)(2x - 2) = 0\]
\[2x - 2 = 0\]
\[x = 1.\]
\[Ответ:x = 0,2.\]
\[3)\ \frac{2x^{3} + 1}{2x + 1} + \frac{3x^{2}}{3x - 1} =\]
\[= \frac{15x^{3}}{6x^{2} + x - 1}\text{\ \ \ \ \ }\]
\[\ | \cdot (2x + 1)(3x - 1)\]
\[6x^{2} + x - 1 =\]
\[= 6 \cdot \left( x + \frac{1}{2} \right)\left( x - \frac{1}{3} \right) =\]
\[= (2x + 1)(3x - 1)\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{- 1 + 5}{12} = \frac{1}{3};\ \ \ \]
\[x_{2} = \frac{- 1 - 5}{12} = - \frac{1}{2}.\]
\[ОДЗ:\ \ x \neq \frac{1}{3};\ \ x \neq \ - \frac{1}{2}.\]
\[\left( 2x^{3} + 1 \right)(3x - 1) +\]
\[+ 3x^{2}(2x + 1) = 15x^{3}\]
\[6x^{4} + 3x - 2x^{3} - 1 + 6x^{3} +\]
\[+ 3x^{2} - 15x^{3} = 0\]
\[6x^{4} - 11x^{3} + 3x^{2} + 3x - 1 = 0\]
\[Делители:\ \pm 1.\]
| \[6\] | \[- 11\] | \[3\] | \[3\] | \[- 1\] | |
|---|---|---|---|---|---|
| \[1\] | \[6\] | \[- 5\] | \[- 2\] | \[1\] | \[0\] |
| \[1\] | \[6\] | \[1\] | \[- 1\] | \[0\] |
\[P(x) =\]
\[= (x - 1)^{2}\left( 6x^{2} + x - 1 \right) = 0\]
\[6x^{2} + x - 1 \neq 0\ (по\ ОДЗ).\]
\[Ответ:x = 1.\]
\[4)\ \frac{6x^{3}}{x + 1} + \frac{5x^{2} - 17x + 2}{x - 2} =\]
\[= \frac{18x}{2 + x - x^{2}}\]
\[\frac{6x^{3}}{x + 1} + \frac{5x^{2} - 17x + 2}{x - 2} =\]
\[= - \frac{18x}{(x + 1)(x - 2)}\text{\ \ \ }\]
\[\ | \cdot (x + 1)(x - 2)\]
\[2 + x - x^{2} = 0\]
\[x^{2} - x - 2 = - (x + 1)(x - 2)\]
\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = 2;\ \ \ x_{2} = - 1.\]
\[ОДЗ:\ \ \text{\ x} \neq - 1;x \neq 2.\]
\[6x^{3}(x - 2) +\]
\[+ \left( 5x^{2} - 17x + 2 \right)(x + 1) +\]
\[+ 18x = 0\]
\[6x^{4} - 12x^{3} + 5x^{3} + 5x^{2} -\]
\[- 17x^{2} - 17x + 2x +\]
\[+ 2 + 18x = 0\]
\[6x^{4} - 7x^{3} - 12x^{2} + 3x + 2 = 0\]
\[Делители:\ \pm 1;\ \pm \frac{1}{2};\ \pm \frac{1}{3}.\]
| \[6\] | \[- 7\] | \[- 12\] | \[3\] | \[2\] | |
|---|---|---|---|---|---|
| \[- 1\] | \[6\] | \[- 13\] | \[1\] | \[2\] | \[0\] |
| \[\frac{1}{2}\] | \[6\] | \[- 10\] | \[- 4\] | \[0\] | |
| \[- \frac{1}{3}\] | \[6\] | \[- 12\] | \[0\] |
\[P(x) = (x + 1)\left( x - \frac{1}{2} \right)\left( x + \frac{1}{3} \right)(6x - 12) = 0.\]
\[6x - 12 = 0\]
\[x = 2\ (не\ подходит\ по\ ОДЗ).\]
\[Ответ:x = - \frac{1}{3};\ \frac{1}{2}.\]
