Решебник по алгебре 9 класс Макарычев Задание 266

 

\[\boxed{\text{266}\text{\ (266)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (8x - 1)(2x - 3) -\]

\[- (4x - 1)^{2} = 38\]

\[16x^{2} - 24x - 2x + 3 - 16x^{2} +\]

\[+ 8x - 1 = 38\]

\[- 18x = 36\]

\[x = - 2.\]

\[Ответ:x = - 2.\]

\[\textbf{б)}\ \frac{(15x - 1)(15x + 1)}{3} = 2\frac{2}{3}\]

\[\frac{225x^{2} - 1}{3} = \frac{8}{3}\]

\[225x^{2} = 9\]

\[x^{2} = \frac{9}{225}\]

\[x = \pm \frac{3}{15} = \pm \frac{1}{5}\]

\[x = \pm 0,2\]

\[Ответ:x = \pm 0,2.\]

\[\textbf{в)}\ 0,5y^{3} - 0,5y(y + 1)(y - 3) =\]

\[= 7\]

\[y^{3} - y\left( y^{2} - 3y + y - 3 \right) = 14\]

\[y^{3} - y^{3} + 2y^{2} + 3y - 14 = 0\]

\[2y^{2} + 3y - 14 = 0\]

\[D = 9 + 4 \cdot 2 \cdot 14 = 121\]

\[y_{1,2} = \frac{- 3 \pm 11}{4} = 2;\ - 3,5.\]

\[Ответ:y = 2;\ \ y = - 3,5.\]

\[\textbf{г)}\ x^{4} - x^{2} = \frac{\left( 1 + 2x^{2} \right)\left( 2x^{2} - 1 \right)}{4}\]

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

\[4x^{2} = 1\]

\[x^{2} = \frac{1}{4}\]

\[x = \pm \frac{1}{2} = \pm 0,5\]

\[Ответ:x = \pm 0,5.\]

\[\boxed{\text{266.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 2x^{2} + 13x - 7 > 0\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\textbf{д)}\ 3x^{2} - 2x > 0\]

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

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

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

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

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

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