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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 316
Задание 316
\[\boxed{\mathbf{316}.}\]
\[1)\ x^{4} + x^{3} - 7x^{2} - x + 6 = 0;\ \]
\[\ x_{1} = 2:\ \]
| \[1\] | \[1\] | \[- 7\] | \[- 1\] | \[6\] | |
|---|---|---|---|---|---|
| \[2\] | \[1\] | \[3\] | \[- 1\] | \[- 3\] | \[0\] |
| \[- 3\] | \[1\] | \[0\] | \[- 1\] | \[0\] |
\[P(x) = (x - 2)(x + 3)\left( x^{2} - 1 \right) =\]
\[= (x - 2)(x + 3)(x + 1)(x - 1) = 0.\]
\[Ответ:x = \pm 1; - 3;\ \ 2.\]
\[2)\ 2x^{4} + 12x^{3} + 11x^{2} + 6x +\]
\[+ 5 = 0;\ \ x_{1} = - 1:\]
| \[2\] | \[12\] | \[11\] | \[6\] | \[5\] | |
|---|---|---|---|---|---|
| \[- 1\] | \[2\] | \[10\] | \[1\] | \[5\] | \[0\] |
| \[- 5\] | \[2\] | \[2\] | \[1\] | \[0\] |
\[P(x) =\]
\[= (x + 1)(x + 5)\left( 2x^{2} + 2x + 1 \right) = 0.\]
\[2x^{2} + 2x + 1 = 0\]
\[D_{1} = 1 - 2 = - 1 < 0\]
\[нет\ корней.\]
\[Ответ:x = - 5;\ - 1.\]
\[3)\ 2x^{5} - x^{4} - 12x^{3} + 6x^{2} +\]
\[+ 18x - 9 = 0;\ \ x_{1} = \frac{1}{2}:\]
\[x^{4}(2x - 1) - 6x^{2}(2x - 1) +\]
\[+ 9 \cdot (2x - 1) = 0\]
\[(2x - 1)\left( x^{4} - 6x^{2} + 9 \right) = 0\]
\[2 \cdot (x - 0,5)\left( x^{2} - 3 \right)^{2} = 0\]
\[x = 0,5;\ \ \ \ \ x^{2} = 3\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \pm \sqrt{3}.\]
\[Ответ:x = 0,5;\ \pm \sqrt{3}.\]
\[4)\ 3x^{5} + x^{4} - 15x^{3} - 5x^{2} +\]
\[+ 12x + 4 = 0;\ \ \ x_{1} = - \frac{1}{3}:\]
\[x^{4}(3x + 1) - 5x^{2}(3x + 1) +\]
\[+ 4 \cdot (3x + 1) = 0\]
\[(3x + 1)\left( x^{4} - 5x^{2} + 4 \right) = 0\]
\[x^{4} - 5x^{2} + 4 = 0\]
\[x^{2} = y \geq 0:\]
\[y^{2} - 5y + 4 = 0\]
\[y_{1} + y_{2} = 5;\ \ y_{1} \cdot y_{2} = 4\]
\[y_{1} = 1;\ \ \ y_{2} = 4.\]
\[x^{2} = 1\ \ \ \ \ \ \ \ \ x^{2} = 4\]
\[x = \pm 1\ \ \ \ \ \ \ \ x = \pm 2.\]
\[3 \cdot \left( x + \frac{1}{3} \right)(x + 2)(x - 2)(x + 1)(x - 1) = 0\]
\[Ответ:x = \pm 2;\ \pm 1;\ - \frac{1}{3}.\]
