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МатематикаРешебник по алгебре 9 класс Макарычев Задание 354
Задание 354
\[\boxed{\text{354\ (354).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x^{3} - x^{2} - 4 \cdot (x - 1)^{2} = 0\]
\[x^{2}(x - 1) - 4 \cdot\]
\[\cdot (x - 1)(x - 1) = 0\]
\[(x - 1)\left( x^{2} - 4 \cdot (x - 1) \right) = 0\]
\[(x - 1)\left( x^{2} - 4x + 4 \right) = 0\]
\[(x - 1)(x - 2)^{2} = 0\]
\[x_{1} = 1;\ \ x_{2} = 2.\]
\[Ответ:x = 1;x = 2.\]
\[\textbf{б)}\ 2y^{3} + 2y^{2} - (y + 1)^{2} = 0\]
\[2y^{2}(y + 1) - (y + 1)(y + 1) = 0\]
\[(y + 1)\left( 2y^{2} - y - 1 \right) = 0\]
\[1)\ y + 1 = 0\ \ \ \]
\[y_{1} = - 1.\]
\[2)\ 2y^{2} - y - 1 = 0\]
\[D = 1 + 4 \cdot 2 = 9\]
\[y_{2,3} = \frac{1 \pm 3}{4} = 1;\ - 0,5.\ \]
\[Ответ:\ y = 0,5;\ \ y = \pm 1.\]
\[\textbf{в)}\ 5x^{3} - 19x^{2} - 38x + 40 = 0\]
\[5x^{3} + 40 - \left( 19x^{2} + 38x \right) = 0\]
\[5 \cdot \left( x^{3} + 8 \right) - 19x(x + 2) = 0\]
\[5 \cdot (x + 2)\left( x^{2} - 2x + 4 \right) -\]
\[- 19x(x + 2) = 0\]
\[(x + 2) \cdot\]
\[\cdot \left( 5 \cdot \left( x^{2} - 2x + 4 \right) - 19x \right) = 0\]
\[(x + 2)\left( 5x^{2} - 29x + 20 \right) = 0\]
\[1)\ x + 2 = 0\ \ \ \ \ \]
\[x_{1} = - 2.\]
\[2)\ 5x^{2} - 29x + 20 = 0\]
\[D = 29^{2} - 4 \cdot 5 \cdot 20 = 441\]
\[x_{2,3} = \frac{29 \pm 21}{10} = 5;0,8.\]
\[Ответ:x = - 2;x = 5;x = 0,8.\]
\[\textbf{г)}\ 6x^{3} - 31x^{2} - 31x + 6 = 0\]
\[6 \cdot \left( x^{3} + 1 \right) - 31x(x + 1) = 0\]
\[6 \cdot (x + 1)\left( x^{2} - x + 1 \right) -\]
\[- 31x(x + 1) = 0\]
\[(x + 1) \cdot\]
\[\cdot \left( 6 \cdot \left( x^{2} - x + 1 \right) - 31x \right) = 0\]
\[(x + 1)\left( 6x^{2} - 6x + 6 - 31x \right) = 0\]
\[(x + 1)\left( 6x^{2} - 37x + 6 \right) = 0\]
\[1)\ x + 1 = 0\ \]
\[x_{1} = - 1.\]
\[2)\ 6x^{2} - 37x + 6 = 0\]
\[D = 37^{2} - 4 \cdot 6 \cdot 6 =\]
\[= 1369 - 144 = 1225\]
\[x_{2,3} = \frac{37 \pm 35}{12} = 6;\frac{1}{6}.\]
\[Ответ:x = - 1;\ \ x = \frac{1}{6};\ \ x = 6.\]
\[\boxed{\text{354.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ y = \frac{4}{\sqrt{(3x - 1)(6x + 1)}}\]
\[(3x - 1)(6x + 1) > 0\]
\[6 \cdot \left( x + \frac{1}{6} \right)\left( x - \frac{1}{3} \right) > 0\]
\[x \in \left( - \infty; - \frac{1}{6} \right) \cup \left( \frac{1}{3}; + \infty \right).\]
\[\textbf{б)}\ y = \frac{7}{\sqrt{(11x + 2)(x - 4)}}\]
\[(11x + 2)(x - 4) > 0\]
\[11 \cdot \left( x + \frac{2}{11} \right)(x - 4) > 0\]
\[x \in \left( - \infty; - \frac{2}{11} \right) \cup (4; + \infty)\text{.\ }\]
