Решебник по алгебре 9 класс Мерзляк Задание 138

 

\[\boxed{\text{138\ (138).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[x^{\backslash 12} - \frac{x + 7^{\backslash 3}}{4} -\]

\[- \frac{11x + 30}{12} < \frac{x - 5^{\backslash 4}}{3}\]

\[\frac{12x - 3x - 21 - 11x - 30}{12} <\]

\[< \frac{4x - 20}{12}\ \ \ \ \ \ \ \ \ \ \ | \cdot 12\]

\[- 2x - 51 < 4x - 20\]

\[- 6x < 31\]

\[x > - 5\frac{1}{6}\text{\ \ }\]

\[x \in \left( - 5\frac{1}{6};\ + \infty \right)\]

\[x = \left\{ - 5; - 4; - 3; - 2; - 1 \right\}.\]

\[Ответ:5.\]

\[\boxed{\mathbf{138.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ x^{4} - x^{2} = 0\]

\[x^{2}\left( x^{2} - 1 \right) = 0\]

\[x^{2}(x - 1)(x + 1) = 0\]

\[x = 0;\ \ x = \pm 1.\]

\[2)\ x^{3} - 100x = 0\]

\[x\left( x^{2} - 100 \right) = 0\]

\[x(x - 10)(x + 10) = 0\]

\[x = 0;\ \ x = \pm 10.\]

\[3)\ 7x^{3} - 14x = 0\]

\[7x\left( x^{2} - 2 \right) = 0\]

\[7x\left( x - \sqrt{2} \right)\left( x + \sqrt{2} \right) = 0\]

\[x = 0;\ \ \ x = \pm \sqrt{2}.\]

\[4)\ 64x^{3} + 16x^{2} + x = 0\]

\[x\left( 64x^{2} + 16x + 1 \right) = 0\]

\[x(8x + 1)^{2} = 0\]

\[64x\left( x + \frac{1}{8} \right)^{2} = 0\]

\[x = 0;\ \ x = - \frac{1}{8}.\]