ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 268

\[\boxed{\text{269.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 0,01x² \leq 1\]

\[x^{2} \leq 100\]

\[x^{2} - 100 \leq 0\]

\[(x - 1)(x + 10) \leq 0\]

\[x \in \lbrack - 10;10\rbrack.\]

\[\textbf{б)}\frac{1}{2}x² > 12\]

\[x² > 24\ \]

\[x² - 24 > 0\]

\[\left( x - 2\sqrt{6} \right)\left( x + 2\sqrt{6} \right) > 0\]

\[x \in \left( - \infty;\ - 2\sqrt{6} \right) \cup \left( 2\sqrt{6}; + \infty \right).\]

\[\textbf{в)}\ 4 \leq - x²\]

\[x^{2} + 4x \leq 0\]

\[x(x + 4) \leq 0\]

\[x \in \lbrack - 4;0\rbrack.\]

\[\textbf{г)}\frac{1}{3}x² > \frac{1}{9}\]

\[x^{2} > \frac{1}{3}\]

\[x^{2} - \frac{1}{3} > 0\]

\[\left( x - \sqrt{\frac{1}{3}} \right)\left( x + \sqrt{\frac{1}{3}} \right) > 0\]

\[x \in \left( - \infty;\ - \frac{\sqrt{3}}{3} \right) \cup \left( \frac{\sqrt{3}}{3}; + \infty \right).\]

\[\textbf{д)}\ 5x² > 2x\]

\[5x^{2} - 2x > 0\]

\[x(5x - 2) > 0\]

\[x \in ( - \infty;0) \cup (0,4; + \infty).\]

\[\textbf{е)} - 0,3x < 0,6x²\]

\[0,6x^{2} + 0,3x > 0\]

\[0,3x(2x + 1) > 0\]

\[x \in ( - \infty;\ - 0,5) \cup (0; + \infty).\]