ГДЗ по алгебре 9 класс Макарычев Задание 306

\[\boxed{\text{306\ (306).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2x^{2} + 13x - 7 > 0\]

\[2x^{2} + 13x - 7 = 0\]

\[D = 169 + 4 \cdot 2 \cdot 7 = 225\]

\[x = \frac{- 13 \pm 15}{4} = 0,5; - 7;\]

\[2 \cdot (x - 0,5)(x + 7) > 0\]

\[x \in ( - \infty;\ - 7) \cup (0,5; + \infty).\]

\[\textbf{б)} - 9x^{2} + 12x - 4 < 0\ |\ :( - 1)\]

\[9x^{2} - 12x + 4 > 0\]

\[9x^{2} - 12x + 4 = 0\ \]

\[(3x - 2)^{2} > 0\]

\[9 \cdot \left( x - \frac{2}{3} \right)^{2} > 0\]

\[x \in \left( - \infty;\frac{2}{3} \right) \cup \left( \frac{2}{3}; + \infty \right).\]

\[\textbf{в)}\ 6x^{2} - 13x + 5 \leq 0\]

\[6x^{2} - 13x + 5 = 0\]

\[D = 169 - 4 \cdot 6 \cdot 5 = 49\]

\[x_{1,2} = \frac{13 \pm 7}{12} = 0,5;1\frac{2}{3};\ \]

\[6 \cdot (x - 0,5)\left( x - 1\frac{2}{3} \right) \leq 0\]

\[x \in \left\lbrack 0,5;1\frac{2}{3} \right\rbrack.\]

\[\textbf{г)} - 2x^{2} - 5x + 18 \leq 0\ |\ :( - 1)\]

\[2x^{2} + 5x - 18 \geq 0\]

\[D = 25 + 4 \cdot 2 \cdot 18 = 169\]

\[x_{1,2} = \frac{- 5 \pm 13}{4} = 2;\ - 4,5;\]

\[2 \cdot (x - 2)(x + 4,5) \geq 0\]

\[x \in ( - \infty; - 4,5\rbrack \cup \lbrack 2; + \infty).\]

\[\textbf{д)}\ 3x^{2} - 2x > 0\]

\[3x\left( x - \frac{2}{3} \right) > 0\]

\[x \in ( - \infty;0) \cup \left( \frac{2}{3}; + \infty \right).\]

\[\textbf{е)}\ 8 - x^{2} < 0\ \ \ \ \ |\ :( - 1)\]

\[x^{2} - 8 > 0\ \]

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

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