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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 352
Задание 352
\[\boxed{\text{352\ (352).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x^{5} - x^{3} = 0\]
\[x^{3}\left( x^{2} - 1 \right) = 0\]
\[x^{3}(x - 1)(x + 1) = 0\]
\[x_{1} = 0;\ \ x_{2} = 1;\ \ x_{3} = - 1.\]
\[Ответ:x = 0;x = \pm 1.\]
\[\textbf{б)}\ x^{6} = 4x^{4}\]
\[x^{6} - 4x^{4} = 0\]
\[x^{4} \cdot \left( x^{2} - 4 \right) = 0\]
\[x^{4} \cdot (x - 2)(x + 2) = 0\]
\[x_{1} = 0;\ \ x_{2} = 2;\ \ x_{3} = - 2.\]
\[Ответ:x = 0;\ \ x = \pm 2.\]
\[\textbf{в)}\ 0,5x^{3} = 32x\ \ \ \ | \cdot 2\]
\[x^{3} = 64x\]
\[x^{3} - 64x = 0\]
\[x\left( x^{2} - 64 \right) = 0\]
\[x(x - 8)(x + 8) = 0\]
\[x_{1} = 0;\ \ x_{2} = 8;\ \ x_{3} = - 8.\]
\[Ответ:x = 0;\ \ x = \pm 8.\]
\[\textbf{г)}\ 0,2x^{4} = 4x^{2}\ \ \ \ \ \ \ \ | \cdot 5\]
\[x^{4} = 20x^{2}\]
\[x^{4} - 20x^{2} = 0\]
\[x^{2}\left( x^{2} - 20 \right) = 0\]
\[x^{2}\left( x - \sqrt{20} \right)\left( x + \sqrt{20} \right) = 0\]
\[x_{1} = 0;\ \ x_{2} = \sqrt{20} = 2\sqrt{5};\ \]
\[\ x_{3} = - 2\sqrt{5}.\]
\[Ответ:x = 0;\ \ x = \pm 2\sqrt{5}.\]
\[\boxed{\text{352.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left( x^{2} - 16 \right)(x + 17) > 0\]
\[(x + 17)(x + 4)(x - 4) > 0\]
\[x \in ( - 17; - 4) \cup (4; + \infty).\]
\[\textbf{б)}\ \left( x - \frac{2}{3} \right)\left( x^{2} - 121 \right) < 0\]
\[(x - 11)\left( x - \frac{2}{3} \right)(x + 11) < 0\]
\[x \in ( - \infty;\ - 11) \cup \left( \frac{2}{3};11 \right).\]
\[\textbf{в)}\ x^{3} - 25x < 0\]
\[x\left( x^{2} - 25 \right) < 0\]
\[(x + 5)x(x - 5) < 0\]
\[x \in ( - \infty; - 5) \cup (0;5).\]
\[\textbf{г)}\ x^{3} - 0,01x > 0\]
\[x\left( x^{2} - 0,01 \right) > 0\]
\[(x + 0,1)x(x - 0,1) > 0\]
\[x \in ( - 0,1;0) \cup (0,1;\ + \infty).\]
\[\textbf{д)}\ \left( x^{2} - 9 \right)\left( x^{2} - 1 \right) > 0\]
\[(x - 3)(x + 3)(x - 1)(x + 1) > 0\]
\[(x + 3)(x + 1)(x - 1)(x - 3) > 0\]
\[x \in ( - \infty; - 3) \cup ( - 1;1) \cup (3; + \infty).\]
\[\textbf{е)}\ \ \left( x^{2} - 15x \right)\left( x^{2} - 36 \right) < 0\]
\[x(x - 15)(x - 6)(x + 6) < 0\]
\[(x + 6)x(x - 6)(x - 15) < 0\]
\[x \in ( - 6;0) \cup (6;15).\]
