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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 320
Задание 320
\[\boxed{\mathbf{320}.}\]
\[1)\ x^{2}(x - 2)(6x + 1) +\]
\[+ x(5x + 3) = 1\]
\[x^{2}\left( 6x^{2} - 12x + x - 2 \right) + 5x^{2} +\]
\[+ 3x - 1 = 0\]
\[6x^{4} - 11x^{3} - 2x^{2} + 5x^{2} + 3x -\]
\[- 1 = 0\]
\[6x^{4} - 11x^{3} + 3x^{2} - 3x - 1\]
| \[6\] | \[- 11\] | \[3\] | \[3\] | \[- 1\] | |
|---|---|---|---|---|---|
| \[1\] | \[6\] | \[- 5\] | \[- 2\] | \[1\] | \[0\] |
| \[1\] | \[6\] | \[1\] | \[- 1\] | \[0\] |
\[P(x) = (x - 1)^{2}\left( 6x^{2} + x - 1 \right) = 0\]
\[6x^{2} + x - 1 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1} = \frac{- 1 + 5}{12} = \frac{4}{12} = \frac{1}{3};\ \ \]
\[x_{2} = \frac{- 1 - 5}{12} = - \frac{1}{2}.\]
\[Ответ:x = - \frac{1}{2};\ \ \frac{1}{3};\ \ 1.\]
\[2)\ x^{2}(3x + 1) - \left( x^{2} + 1 \right)^{2} = 3\]
\[3x^{3} + x^{2} - x^{4} - 2x^{2} -\]
\[- 1 - 3 = 0\]
\[x^{4} - 3x^{3} + x^{2} + 4 = 0\]
\[Делители:\ \pm 1;\ \pm 2;\ \pm 3.\]
| \[1\] | \[- 3\] | \[1\] | \[0\] | \[4\] | |
|---|---|---|---|---|---|
| \[2\] | \[1\] | \[- 1\] | \[- 1\] | \[- 2\] | \[0\] |
| \[2\] | \[1\] | \[1\] | \[1\] | \[0\] |
\[P(x) = (x - 2)^{2}\left( x^{2} + x + 1 \right) = 0\]
\[x^{2} + x + 1 = 0\]
\[D = 1 - 4 < 0\]
\[нет\ корней.\]
\[Ответ:x = 2.\]
