Решебник по алгебре 9 класс Макарычев Задание 211

 

\[\boxed{\text{211}\text{\ (211)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = \sqrt{1 - x}\]

\[Так\ как\ график\ функции\ \]

\[убывает,\ область\ определения\ \]

\[( - \infty;1\rbrack.\]

\[\boxed{\text{211.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[- 18x = 36\]

\[x = - 2.\]

\[Ответ:x = - 2.\]

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

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

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

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

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

\[x = \pm 0,2\]

\[Ответ:x = \pm 0,2.\]

\[\textbf{в)}\ 0,5y³ - 0,5y(y + 1)(y - 3) =\]

\[= 7\]

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

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

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

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

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

\[Ответ:y = 2;\ \ y = - 3,5.\]

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

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

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

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

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

\[Ответ:x = \pm 0,5.\]