ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 221

\[\boxed{\text{222.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[Пусть\ x^{2} + 3 = t,\ \ \ тогда\ \]

\[t^{2} - 11t + 28 = 0\]

\[D = 121 - 4 \cdot 28 = 9\]

\[t_{1,2} = \frac{11 \pm 3}{2},\ \ t_{1} = 7,\]

\[\text{\ \ }t_{2} = 4\]

\[\left\{ \begin{matrix} x^{2} + 3 = 7 \\ x^{2} + 3 = 4 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x^{2} = 4 \\ x^{2} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x_{1,2} = \pm 2 \\ x_{3,4} = \pm 1 \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm 2;x = \pm 1.\]

\[Пусть\ x^{2} - 4x = t,\ \ тогда\]

\[\ t^{2} + 9t + 20 = 0\]

\[D = 81 - 80 = 1\]

\[t_{1,2} = \frac{- 9 \pm 1}{2},\ \ t_{1} = - 5,\]

\[\text{\ \ }t_{2} = - 4;\]

\[\left\{ \begin{matrix} x^{2} - 4x = - 5 \\ x^{2} - 4x = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} - 4x + 5 = 0\ \ (1) \\ x^{2} - 4x + 4 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ D = 4 - 5 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет;\]

\[(2)\ D = 4 - 4 = 0,\ \ \]

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

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

\[x = 2\]

\[Ответ:x = 2.\]

\[\textbf{в)}\ \left( x^{2} + 1 \right)\left( x^{2} + x - 5 \right) = 84\]

\[Пусть\ \ x^{2} + x = t,\ \ тогда\ \ \]

\[\ t(t - 5) = 84\]

\[t^{2} - 5t - 84 = 0\]

\[D = 25 + 4 \cdot 84 = 361 = 19^{2}\]

\[t_{1,2} = \frac{5 \pm 19}{2},\ \ t_{1} = 12,\ \ \]

\[t_{2} = - 7;\]

\[\left\{ \begin{matrix} x^{2} + x = 12 \\ x^{2} + x = - 7 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x² + x - 12 = 0\ \ (1) \\ x² + x + 7 = 0\ \ \ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ D = 1 + 4 \cdot 12 = 49,\ \ \]

\[x_{1,2} = \frac{- 1 \pm 7}{2} = - 4;3;\]

\[(2)\ D = 1 - 4 \cdot 7 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:x = 3;\ \ x = - 4.\]