ГДЗ по алгебре 9 класс Макарычев Задание 52

 

\[\boxed{\text{52\ (52).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 0,6x^{2} - 3,6x = 0\]

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

\[x(x - 6) = 0\]

\[x_{1} = 0\ \ \ и\ \ \ x_{2} = 6.\]

\[Ответ:x = 0;x = 6.\]

\[\textbf{б)}\ x^{2} - 5 = 0\]

\[x^{2} = 5\]

\[x_{1,2} = \pm \sqrt{5}.\]

\[Ответ:x = \pm \sqrt{5}.\]

\[\textbf{в)}\ 2x^{2} + 17x = 0\]

\[x(2x + 17) = 0\]

\[x_{1} = 0\]

\[2x_{2} + 17 = 0\]

\[2x_{2} = - 17\]

\[x_{2} = - 8,5.\]

\[Ответ:x = 0;x = - 8,5.\]

\[\textbf{г)}\ 0,5x^{2} + 9 = 0\]

\[0,5x^{2} = - 9\]

\[x^{2} = - 18 \Longrightarrow корней\ нет.\]

\[Ответ:нет\ решений.\]

\[\boxed{\text{52.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 6a(a + 1) <\]

\[< (3a + 1)(2a + 1) + a\]

\[6a^{2} + 6a <\]

\[< 6a^{2} + 3a + 2a + 1 + a\]

\[6a^{2} + 6a - 6a^{2} - 6a < 1\]

\[0 < 1 \Longrightarrow ч.т.д.\]

\[4p^{2} + 2p - 2p - 1 + 3p + 3 >\]

\[> 4p^{2} + 3p\]

\[4p^{2} + 3p - 4p^{2} - 3p > 1 - 3\]

\[0 > - 2\ \ \ | \cdot ( - 1)\]

\[2 > 0 \Longrightarrow ч.т.д.\]