Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 9 класс Макарычев Задание 277
Задание 277
\[\boxed{\text{277}\text{\ (277)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left( x^{2} + 3 \right)^{2} - 11 \cdot \left( x^{2} + 3 \right) +\]
\[+ 28 = 0\]
\[Пусть\ x^{2} + 3 = t:\ \]
\[t^{2} - 11t + 28 = 0\]
\[t_{1} + t_{2} = 11;\ \ \ t_{1} \cdot t_{2} = 28\]
\[\text{\ \ }t_{1} = 7;\ \ \ \ \ t_{2} = 4\]
\[\left\{ \begin{matrix} x^{2} + 3 = 7 \\ x^{2} + 3 = 4 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x^{2} = 4 \\ x^{2} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1,2} = \pm 2 \\ x_{3,4} = \pm 1 \\ \end{matrix} \right.\ \]
\[Ответ:x = \pm 2;x = \pm 1.\]
\[\textbf{б)}\ \left( x^{2} - 4x \right)^{2} + 9 \cdot \left( x^{2} - 4x \right) +\]
\[+ 20 = 0\]
\[Пусть\ x^{2} - 4x = t:\ \]
\[t^{2} + 9t + 20 = 0\]
\[t_{1} + t_{2} = - 9;\ \ \ t_{1} \cdot t_{2} = 20\]
\[t_{1} = - 5;\ \ \ t_{2} = - 4;\]
\[\left\{ \begin{matrix} x^{2} - 4x = - 5 \\ x^{2} - 4x = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 4x + 5 = 0\ \ (1) \\ x^{2} - 4x + 4 = 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ x^{2} - 4x + 5 = 0\]
\[D_{1} = 4 - 5 = - 1 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет;\]
\[(2)\ x^{2} - 4x + 4 = 0\]
\[(x - 2)^{2} = 0\]
\[x - 2 = 0\]
\[x = 2\]
\[Ответ:x = 2.\]
\[\textbf{в)}\ \left( x^{2} + 1 \right)\left( x^{2} + x - 5 \right) = 84\]
\[Пусть\ \ x^{2} + x = t:\]
\[t(t - 5) = 84\]
\[t^{2} - 5t - 84 = 0\]
\[t_{1} + t_{2} = 5;\ \ \ t_{1} \cdot t_{2} = - 84\]
\[t_{1} = 12;\ \ \ \ t_{2} = - 7;\]
\[\left\{ \begin{matrix} x^{2} + x = 12 \\ x^{2} + x = - 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x^{2} + x - 12 = 0\ \ (1) \\ x^{2} + x + 7 = 0\ \ \ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ x^{2} + x - 12 = 0\]
\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 12\]
\[x_{1} = - 4;\ \ \ x_{2} = 3.\]
\[(2)\ x^{2} + x + 7 = 0\]
\[D = 1 - 4 \cdot 7 = - 27 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет.\]
\[Ответ:x = 3;\ \ x = - 4.\]
\[\boxed{\text{277.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ x² + 7x + 1 > - x^{2} + 10x - 1\]
\[2x^{2} - 3x + 2 > 0\]
\[D = 9 - 4 \cdot 2 \cdot 2 < 0\]
\[\Longrightarrow 2x^{2} - 3x + 2 > 0\ \ при\ \]
\[любом\ x;\]
\[\Longrightarrow x^{2} + 7x + 1 >\]
\[> - x^{2} + 10 - 1 \Longrightarrow при\ \]
\[любом\ x.\]
\[\textbf{б)} - 2x^{2} + 10x < 18 - 2x\]
\[2x^{2} - 12x + 18 > 0\]
\[x² - 6x + 9 > 0\ \]
\[(x - 3)^{2} > 0\]
\[x \in ( - \infty;3) \cup (3; + \infty).\]
\[\Longrightarrow - 2x^{2} + 10x < 18 - 2x \Longrightarrow\]
\[\Longrightarrow при\ любом\ x \neq 3.\]
