Решите неравенство методом интервалов: x^3-64x<0.

Ответ:

\[\ x^{3} - 64x < 0\]

\[x\left( x^{2} - 64 \right) < 0\]

\[x = 0\ \ \ \ \ \ \ \ \ \ x^{2} = 64\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \pm 8\]

\[Ответ:x \in ( - \infty;\ \ - 8) \cup (0;8).\]

\[\sqrt{x^{2} - 2x - 35}\]

\[x^{2} - 2x - 35 \geq 0\]

\[x^{2} - 2x - 35 = 0\]

\[x_{1} + x_{2} = 2\]

\[x_{1} \cdot x_{2} = - 35 \Longrightarrow \ \ x_{1} = 7\ \ \ \ и\ \ x_{2} = - 5.\]

\[Ответ:\ \ x\ \in ( - \infty;\ - 5\rbrack\ \cup \lbrack 7;\ + \infty).\]

\[\ 2x² + 5x - 12 > 0\]

\[D = b^{2} - 4ac = 25 - 4 \cdot 2 \cdot ( - 12) =\]

\[= 25 + 96 = 121\]

\[x_{1} = \frac{- 5 + 11}{4} = \frac{6}{4} = 1,5\]

\[x_{2} = \frac{- 5 - 11}{4} = - \frac{16}{4} = - 4\]

\[Ответ:\ \ x \in ( - \infty; - 4) \cup (1,5;\ + \infty).\]