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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 273
Задание 273
\[\boxed{\text{273\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \ 3x^{3} - x^{2} + 18x - 6 = 0\]
\[x^{2}(3x - 1) + 6 \cdot (3x - 1) = 0\]
\[\left( x^{2} + 6 \right)(3x - 1) = 0\]
\[x_{1}^{2} = - 6 \Longrightarrow корней\ нет;\]
\[3x_{2} = 1\ \ \]
\[x_{2} = \frac{1}{3}\]
\[Ответ:x = \frac{1}{3}.\]
\[\textbf{б)}\ 2x^{4} - 18x^{2} = 5x^{3} - 45x\]
\[2x^{2}\left( x^{2} - 9 \right) = 5x\left( x^{2} - 9 \right)\]
\[\left( 2x^{2} - 5x \right)\left( x^{2} - 9 \right) = 0\]
\[x(2x - 5)(x - 3)(x + 3) = 0\]
\[x_{1} = 0,\ \ x_{2} = 2,5;\ \]
\[x_{3} = 3,\ \ x_{4} = - 3;\]
\[Ответ:x = 0;\ \ x = 2,5;\ \ x = \pm 3.\]
\[\boxed{\text{273\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \ 0,7x^{4} - x^{3} = 0\]
\[x^{3}(0,7x - 1) = 0\]
\[x_{1} = 0\ \ \]
\[И\ \]
\[0,7x_{2} = 1\ \ \]
\[x_{2} = \frac{10}{7} = 1\frac{3}{7}\]
\[Ответ:x = 0;\ \ x = 1\frac{3}{7}.\]
\[\textbf{б)}\ 0,5x^{3} - 72x = 0\]
\[x\left( 0,5x^{2} - 72 \right) = 0\]
\[x\left( x^{2} - 144 \right) = 0\]
\[x(x - 12)(x + 12) = 0\]
\[{x_{1} = 0,\ \ x_{2} = 12,\ \ }{x_{3} = - 12;}\]
\[Ответ:x = 0;\ \ x = \pm 12.\]
\[\textbf{в)}\ x^{3} + 4x = 5x^{2}\]
\[x^{3} - 5x^{2} + 4x = 0\]
\[x \cdot \left( x^{2} - 5x + 4 \right) = 0\]
\[x^{2} - 5x + 4 = 0\]
\[D = 25 - 16 = 9\]
\[x_{1,2} = \frac{5 \pm 3}{2} = 4;1;\ \ \ \ \]
\[x_{3} = 0\]
\[Ответ:x = 1;\ \ x = 4;\ \ x = 0.\]
\[\textbf{г)}\ 3x^{3} - x^{2} + 18x - 6 = 0\]
\[x^{2}(3x - 1) + 6 \cdot (3x - 1) = 0\]
\[\left( x^{2} + 6 \right)(3x - 1) = 0\]
\[x_{1}^{2} = - 6 \Longrightarrow корней\ нет;\]
\[3x_{2} = 1\ \ \]
\[x_{2} = \frac{1}{3}\]
\[Ответ:x = \frac{1}{3}.\]
\[\textbf{д)}\ 2x^{4} - 18x^{2} = 5x^{3} - 45x\]
\[2x^{2}\left( x^{2} - 9 \right) = 5x\left( x^{2} - 9 \right)\]
\[\left( 2x^{2} - 5x \right)\left( x^{2} - 9 \right) = 0\]
\[x(2x - 5)(x - 3)(x + 3) = 0\]
\[x_{1} = 0,\ \ x_{2} = 2,5;\ \ \]
\[\ x_{3} = 3,\ \ x_{4} = - 3;\]
\[Ответ:x = 0;\ \ x = 2,5;\ \ x = \pm 3.\]
\[\textbf{е)}\ 3y^{2} - 2y = 2y^{3} - 3\]
\[2y^{3} - 3y^{2} + 2y - 3 = 0\ \]
\[2y\left( y^{2} + 1 \right) - 3 \cdot \left( y^{2} + 1 \right) = 0\]
\[(2y - 3)\left( y^{2} + 1 \right) = 0\]
\[y_{1} = 1,5;\ \ \ \]
\[y_{2}^{2} = - 1 \Longrightarrow нет\ корней.\]
\[Ответ:y = 1,5.\]
