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МатематикаРешебник по алгебре 9 класс Макарычев Задание 273
Задание 273
\[\boxed{\text{273\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \ 3x^{3} - x^{2} + 18x - 6 = 0\]
\[x^{2}(3x - 1) + 6 \cdot (3x - 1) = 0\]
\[\left( x^{2} + 6 \right)(3x - 1) = 0\]
\[x_{1}^{2} = - 6 \Longrightarrow корней\ нет;\]
\[3x_{2} = 1\ \ \]
\[x_{2} = \frac{1}{3}\]
\[Ответ:x = \frac{1}{3}.\]
\[\textbf{б)}\ 2x^{4} - 18x^{2} = 5x^{3} - 45x\]
\[2x^{2}\left( x^{2} - 9 \right) = 5x\left( x^{2} - 9 \right)\]
\[\left( 2x^{2} - 5x \right)\left( x^{2} - 9 \right) = 0\]
\[x(2x - 5)(x - 3)(x + 3) = 0\]
\[x_{1} = 0,\ \ x_{2} = 2,5;\ \]
\[x_{3} = 3,\ \ x_{4} = - 3;\]
\[Ответ:x = 0;\ \ x = 2,5;\ \ x = \pm 3.\]
\[\boxed{\text{273\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \ 0,7x^{4} - x^{3} = 0\]
\[x^{3}(0,7x - 1) = 0\]
\[x_{1} = 0\ \ \]
\[И\ \]
\[0,7x_{2} = 1\ \ \]
\[x_{2} = \frac{10}{7} = 1\frac{3}{7}\]
\[Ответ:x = 0;\ \ x = 1\frac{3}{7}.\]
\[\textbf{б)}\ 0,5x^{3} - 72x = 0\]
\[x\left( 0,5x^{2} - 72 \right) = 0\]
\[x\left( x^{2} - 144 \right) = 0\]
\[x(x - 12)(x + 12) = 0\]
\[{x_{1} = 0,\ \ x_{2} = 12,\ \ }{x_{3} = - 12;}\]
\[Ответ:x = 0;\ \ x = \pm 12.\]
\[\textbf{в)}\ x^{3} + 4x = 5x^{2}\]
\[x^{3} - 5x^{2} + 4x = 0\]
\[x \cdot \left( x^{2} - 5x + 4 \right) = 0\]
\[x^{2} - 5x + 4 = 0\]
\[D = 25 - 16 = 9\]
\[x_{1,2} = \frac{5 \pm 3}{2} = 4;1;\ \ \ \ \]
\[x_{3} = 0\]
\[Ответ:x = 1;\ \ x = 4;\ \ x = 0.\]
\[\textbf{г)}\ 3x^{3} - x^{2} + 18x - 6 = 0\]
\[x^{2}(3x - 1) + 6 \cdot (3x - 1) = 0\]
\[\left( x^{2} + 6 \right)(3x - 1) = 0\]
\[x_{1}^{2} = - 6 \Longrightarrow корней\ нет;\]
\[3x_{2} = 1\ \ \]
\[x_{2} = \frac{1}{3}\]
\[Ответ:x = \frac{1}{3}.\]
\[\textbf{д)}\ 2x^{4} - 18x^{2} = 5x^{3} - 45x\]
\[2x^{2}\left( x^{2} - 9 \right) = 5x\left( x^{2} - 9 \right)\]
\[\left( 2x^{2} - 5x \right)\left( x^{2} - 9 \right) = 0\]
\[x(2x - 5)(x - 3)(x + 3) = 0\]
\[x_{1} = 0,\ \ x_{2} = 2,5;\ \ \]
\[\ x_{3} = 3,\ \ x_{4} = - 3;\]
\[Ответ:x = 0;\ \ x = 2,5;\ \ x = \pm 3.\]
\[\textbf{е)}\ 3y^{2} - 2y = 2y^{3} - 3\]
\[2y^{3} - 3y^{2} + 2y - 3 = 0\ \]
\[2y\left( y^{2} + 1 \right) - 3 \cdot \left( y^{2} + 1 \right) = 0\]
\[(2y - 3)\left( y^{2} + 1 \right) = 0\]
\[y_{1} = 1,5;\ \ \ \]
\[y_{2}^{2} = - 1 \Longrightarrow нет\ корней.\]
\[Ответ:y = 1,5.\]
\[\boxed{\text{273.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ 2x(3x - 1) > 4x² + 5x + 9\]
\[6x^{2} - 2x > 4x^{2} + 5x + 9\]
\[2x^{2} - 7x - 9 > 0\]
\[D = 49 + 4 \cdot 2 \cdot 9 = 121\]
\[x_{1,2} = \frac{7 \pm 11}{4},\ \ x_{1} = - 1,\]
\[\text{\ \ }x_{2} = 4,5.\]
\[2 \cdot (x + 1)(x - 4,5) > 0\]
\[x \in ( - \infty;\ - 1) \cup (4,5;\ + \infty).\]
\[\textbf{б)}(5x + 7)(x - 2) <\]
\[< 21x² - 11x - 13\]
\[5x^{2} - 10x + 7x - 14 <\]
\[< 21x^{2} - 11x - 13\]
\[16x^{2} - 8x + 1 > 0\]
\[(4x - 1)² > 0\ \]
\[x \in ( - \infty;0,25) \cup (0,25;\ + \infty).\]
