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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 390
Задание 390
\[\boxed{\mathbf{390}.}\]
\[1)\ \left( x^{2} + 2x \right)^{2} - 2x^{2} -\]
\[- 4x - 3 = 0\]
\[x^{4} + 4x^{3} + 4x^{2} - 2x^{2} -\]
\[- 4x - 3 = 0\]
\[x^{4} + 4x^{3} + 2x^{2} - 4x - 3 = 0\]
\[Делители:\ \pm 1;\ \pm 3.\]
| \[1\] | \[4\] | \[2\] | \[- 4\] | \[- 3\] | |
|---|---|---|---|---|---|
| \[1\] | \[1\] | \[5\] | \[7\] | \[3\] | \[0\] |
| \[- 1\] | \[1\] | \[4\] | \[3\] | \[0\] | |
| \[- 1\] | \[1\] | \[3\] | \[0\] |
\[P(x) =\]
\[= (x - 1)(x + 1)^{2}(x - 3) = 0.\]
\[Ответ:x = \pm 1;\ \ 3.\]
\[2)\ \left( x^{2} - x - 3 \right)\left( x^{2} - x - 2 \right) =\]
\[= 12\]
\[t = x^{2} - x:\]
\[(t - 3)(t - 2) = 12\]
\[t^{2} - 3t - 2t + 6 - 12 = 0\]
\[t^{2} - 5t - 6 = 0\]
\[t_{1} + t_{2} = 5;\ \ \ t_{1} \cdot t_{2} = - 6\]
\[t_{1} = - 1;\ \ t_{2} = 6.\]
\[1)\ x^{2} - x = - 1\]
\[x^{2} - x + 1 = 0\]
\[D = 1 - 4 < 0\]
\[нет\ корней.\]
\[2)\ x^{2} - x = 6\]
\[x^{2} - x - 6 = 0\]
\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]
\[x_{1} = 3;\ \ \ x_{2} = - 2.\]
\[Ответ:x = - 2;\ \ 3.\]
\[3)\ \left( x^{2} + x \right)^{2} + (3x - 1)x^{2} +\]
\[+ 5x(x - 1) = 6\]
\[x^{4} + 2x^{3} + x^{2} + 3x^{3} - x^{2} +\]
\[+ 5x^{2} - 5x - 6 = 0\]
\[x^{4} + 5x^{3} + 5x^{2} - 5x - 6 = 0\]
\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 6.\]
| \[1\] | \[5\] | \[5\] | \[- 5\] | \[- 6\] | |
|---|---|---|---|---|---|
| \[1\] | \[1\] | \[6\] | \[11\] | \[6\] | \[0\] |
| \[- 1\] | \[1\] | \[5\] | \[6\] | \[0\] | |
| \[- 2\] | \[1\] | \[3\] | \[0\] |
\[P(x) = (x - 1)(x + 1)\]
\[(x + 2)(x + 3) = 0.\]
\[Ответ:x = - 3;\ - 2;\ \pm 1.\]
\[4)\ x^{2}\left( x^{2} - 5 \right) - 2x\left( x^{2} - 4 \right) +\]
\[+ 4 = 0\]
\[x^{4} - 5x^{2} - 2x^{3} + 8x + 4 = 0\]
\[x^{4} - 2x^{3} - 5x^{2} + 8x + 4 = 0\]
\[Делители:\ \pm 1; \pm 2; \pm 4.\]
| \[1\] | \[- 2\] | \[- 5\] | \[8\] | \[4\] | |
|---|---|---|---|---|---|
| \[2\] | \[1\] | \[0\] | \[- 5\] | \[2\] | \[0\] |
| \[- 2\] | \[1\] | \[- 2\] | \[- 1\] | \[0\] |
\[P(x) = (x - 2)(x + 2)\]
\[\left( x^{2} - 2x - 1 \right) = 0\]
\[x^{2} - 2x - 1 = 0\]
\[D_{1} = 1 + 1 = 2\]
\[x = \pm \sqrt{2}.\]
\[Ответ:x = \pm 2;\ \pm \sqrt{2}.\]
\[5)\ \left( x^{2} - 2x \right)^{2} - 4x\left( x^{2} + 2 \right) +\]
\[+ 4 \cdot (10x - 1) = 7x^{2}\]
\[x^{4} - 4x^{3} + 4x^{2} - 4x^{3} - 8x +\]
\[+ 40x - 4 - 7x^{2} = 0\]
\[x^{4} - 8x^{3} - 3x^{2} + 32x - 4 = 0\]
\[Делители:\ \pm 1;\ \pm 2;\ \pm 4.\]
| \[1\] | \[- 8\] | \[- 3\] | \[32\] | \[- 4\] | |
|---|---|---|---|---|---|
| \[2\] | \[1\] | \[- 6\] | \[- 15\] | \[2\] | \[0\] |
| \[- 2\] | \[1\] | \[- 8\] | \[1\] | \[0\] |
\[P(x) = (x - 2)(x + 2)\]
\[\left( x^{2} - 8x + 1 \right) = 0\]
\[x^{2} - 8x + 1 = 0\]
\[D_{1} = 16 - 1 = 15\]
\[x = 4 \pm \sqrt{15}.\]
\[Ответ:x = \pm 2;\ 4 \pm \sqrt{15}.\]
\[6)\ \left( x^{2} - 2 \right)^{2} +\]
\[+ x(x - 1)(x + 1) = 1\]
\[x^{4} - 4x^{2} + 4 + x^{3} - x - 1 = 0\]
\[x^{4} + x^{3} - 4x^{2} - x + 3 = 0\]
\[Делители:\ \pm 1;\ \pm 3.\]
| \[1\] | \[1\] | \[- 4\] | \[- 1\] | \[3\] | |
|---|---|---|---|---|---|
| \[1\] | \[1\] | \[2\] | \[- 2\] | \[- 3\] | \[0\] |
| \[- 1\] | \[1\] | \[1\] | \[- 3\] | \[0\] |
\[P(x) = (x - 1)(x + 1)\]
\[\left( x^{2} + x - 3 \right) = 0\]
\[x^{2} + x - 3 = 0\]
\[D = 1 + 12 = 13\]
\[x = \frac{- 1 \pm \sqrt{13}}{2}.\]
\[Ответ:x = \pm 1;\ \frac{- 1 \pm \sqrt{13}}{2}\text{.\ }\]
