Решебник по алгебре 9 класс Макарычев Задание 267

 

\[\boxed{\text{267}\text{\ (267)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (6 - x)(x + 6) -\]

\[- (x - 11) \cdot x = 36\]

\[36 - x^{2} - x^{2} + 11x - 36 = 0\]

\[- 2x^{2} + 11x = 0\]

\[x(11 - 2x) = 0\]

\[x_{1} = 0,\ \ x_{2} = \frac{11}{2} = 5,5\]

\[Ответ:x = 0;x = 5,5.\]

\[\textbf{б)}\ \frac{1 - 3y}{11} - \frac{3 - y}{5} = 0\ \ \ \ \ \ | \cdot 55\]

\[5 - 15y - 33 + 11y = 0\]

\[4y = - 28\]

\[y = - 7\]

\[Ответ:y = - 7.\]

\[\textbf{в)}\ 9x^{2} - \frac{(12x - 11)(3x + 8)}{4} =\]

\[= 1\ \ \ \ \ \ \ | \cdot 4\]

\[36x^{2} -\]

\[- \left( 36x^{2} + 96x - 33x - 88 \right) = 4\]

\[- 63x = - 84\]

\[x = \frac{84}{63}\]

\[x = 1\frac{1}{3}.\]

\[Ответ:x = 1\frac{1}{3}.\]

\[\textbf{г)}\ \frac{(y + 1)^{2}}{12} - \frac{1 - y^{2}}{24} = 4\ \ | \cdot 24\]

\[2 \cdot \left( y^{2} + 2y + 1 \right) - 1 + y^{2} = 96\]

\[2y^{2} + 4y + 2 - 1 + y^{2} - 96 = 0\]

\[3y^{2} + 4y - 95 = 0\]

\[D = 2^{2} + 3 \cdot 95 = 289\]

\[y_{1,2} = \frac{- 2 \pm 17}{3},\ \ \]

\[y_{1} = 5,\ \ y_{2} = - 6\frac{1}{3}.\]

\[Ответ:y = 5;\ \ y = - 6\frac{1}{3}.\]

\[\boxed{\text{267.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 2x^{2} + 5x + 3 > 0\]

\[2x^{2} + 5x + 3 = 0\]

\[D = 25 - 4 \cdot 2 \cdot 3 = 1\]

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

\[2 \cdot (x + 1,5)(x + 1) > 0\]

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

\[\textbf{б)} - x^{2} - \frac{1}{3}x - \frac{1}{36} < 0\]

\[x^{2} + \frac{1}{3}x + \frac{1}{36} > 0\ \ \ \ \ \ \ \ \ \ | \cdot 36\ \]

\[36x^{2} + 12x + 1 > 0\]

\[(6x + 1)^{2} > 0\]

\[36 \cdot \left( x + \frac{1}{6} \right)^{2} > 0\]

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