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МатематикаРешебник по алгебре 9 класс Макарычев Задание 289
Задание 289
\[\boxed{\text{289\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{5y^{3} - 15y^{2} - 2y + 6}{y^{2} - 9} = 0\]
\[\frac{5y^{2}(y - 3) - 2 \cdot (y - 3)}{y^{2} - 9} = 0\]
\[\frac{\left( 5y^{2} - 2 \right)(y - 3)}{(y - 3)(y + 3)} = 0\]
\[\frac{5y^{2} - 2}{y + 3} = 0\]
\[5y^{2} = 2\]
\[y^{2} = \frac{2}{5}\]
\[y = \pm \sqrt{\frac{2}{5}}.\]
\[Ответ:y = \pm \sqrt{\frac{2}{5}}.\]
\[\textbf{б)}\ \frac{3y^{3} - 12y^{2} - y + 4}{9y^{4} - 1} = 0\]
\[\frac{3y^{2}(y - 4) - (y - 4)}{9y^{4} - 1} = 0\]
\[\frac{(y - 4)\left( 3y^{2} - 1 \right)}{\left( 3y^{2} - 1 \right)\left( 3y^{2} + 1 \right)} = 0\]
\[\frac{y - 4}{3y^{2} + 1} = 0\]
\[y - 4 = 0\]
\[y = 4.\]
\[Ответ:y = 4.\]
\[\textbf{в)}\ \frac{6x^{3} + 48x^{2} - 2x - 16}{x^{2} - 64} = 0\]
\[\frac{6x^{2}(x + 8) - 2 \cdot (x + 8)}{x^{2} - 64} = 0\]
\[\frac{\left( 6x^{2} - 2 \right)(x + 8)}{(x - 8)(x + 8)} = 0\]
\[\frac{6x^{2} - 2}{x - 8} = 0\]
\[6x^{2} - 2 = 0\]
\[6x^{2} = 2\]
\[x^{2} = \frac{1}{3}\]
\[x = \pm \sqrt{\frac{1}{3}}\]
\[Ответ:x = \pm \sqrt{\frac{1}{3}}.\]
\[\textbf{г)}\ \frac{y^{3} - 4y^{2} - 6y + 24}{y^{3} - 6y} = 0\]
\[\frac{y^{2}(y - 4) - 6 \cdot (y - 4)}{y\left( y^{2} - 6 \right)} = 0\]
\[\frac{\left( y^{2} - 6 \right)(y - 4)}{y\left( y^{2} - 6 \right)} = 0\]
\[\frac{y - 4}{y} = 0\]
\[y - 4 = 0\]
\[y = 4\]
\[Ответ:y = 4.\]
\[\boxed{\text{289\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{5y^{3} - 15y^{2} - 2y + 6}{y^{2} - 9} = 0\]
\[\frac{5y^{2}(y - 3) - 2 \cdot (y - 3)}{y^{2} - 9} = 0\]
\[\frac{\left( 5y^{2} - 2 \right)(y - 3)}{(y - 3)(y + 3)} = 0\]
\[\frac{5y^{2} - 2}{y + 3} = 0\]
\[5y^{2} = 2\]
\[y^{2} = \frac{2}{5}\]
\[y = \pm \sqrt{\frac{2}{5}}.\]
\[Ответ:y = \pm \sqrt{\frac{2}{5}}.\]
\[\textbf{б)}\ \frac{3y^{3} - 12y^{2} - y + 4}{9y^{4} - 1} = 0\]
\[\frac{3y^{2}(y - 4) - (y - 4)}{9y^{4} - 1} = 0\]
\[\frac{(y - 4)\left( 3y^{2} - 1 \right)}{\left( 3y^{2} - 1 \right)\left( 3y^{2} + 1 \right)} = 0\]
\[\frac{y - 4}{3y^{2} + 1} = 0\]
\[y - 4 = 0\]
\[y = 4.\]
\[Ответ:y = 4.\]
\[\boxed{\text{289.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[\textbf{а)}\ (x + 9)(x - 2)(x - 15) < 0\]
\[x \in ( - \infty;\ - 9) \cup (2;15).\]
\[\textbf{б)}\ x(x - 5)(x + 6) > 0\]
\[(x + 6)x(x - 5) > 0\]
\[x \in ( - 6;0) \cup (5;\ + \infty).\]
\[x \in (1;4) \cup (8;16)\text{.\ }\]
