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МатематикаГДЗ по алгебре 9 класс Макарычев ФГОС Задание 220
Задание 220
\[\boxed{\text{221.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Пусть\ 2x^{2} + 3 = t,\ \ тогда\ \]
\[t^{2} - 12t + 11 = 0,\]
\[D = 36 - 11 = 25\]
\[t_{1,2} = 6 \pm 5 = 11;1:\]
\[\left\{ \begin{matrix} 2x^{2} + 3 = 11 \\ 2x^{2} + 3 = 1\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 2x^{2} = 8 \\ 2x^{2} = - 1 \\ \end{matrix}\ \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x² = 4\ \ \ \\ x² = - 1 \\ \end{matrix} \right.\ \]
\[x^{2} = 4 \Longrightarrow x_{1,2} = \pm 2,\]
\[x² = - 1 \Longrightarrow корней\ не\ имеет.\]
\[Ответ:x = \pm 2.\]
\[\textbf{б)}\ \left( t^{2} - 2t \right)^{2} - 3 = 2 \cdot (t^{2} - 2t)\]
\[Пусть\ \ t^{2} - 2t = y,\ \ тогда\ \ \]
\[y^{2} - 3 = 2y\]
\[y^{2} - 2y - 3 = 0\]
\[D = 1 + 3 = 4\]
\[y_{1,2} = 1 \pm 2 = 3;\ - 1:\]
\[\left\{ \begin{matrix} t^{2} - 2t = 3\ \ \\ t^{2} - 2t = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} t² - 2t - 3 = 0\ \ (1) \\ t² - 2t + 1 = 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\text{\ \ }D = 1 + 3 = 4,\ \ \]
\[t_{1,2} = 1 \pm 2 = 3;\ - 1;\]
\[(2)\ (t - 1)^{2} = 0,\ \ \]
\[t_{3} = 1.\]
\[Ответ:t = 3;t = \pm 1.\]
\[\textbf{в)}\ \left( x^{2} + x - 1 \right)\left( x^{2} + x + 2 \right) =\]
\[= 40\]
\[Пусть\ x^{2} + x - 1 = a,\ \ тогда\ \]
\[\ a \cdot (a + 3) = 40\]
\[a^{2} + 3a - 40 = 0\]
\[D = 9 + 160 = 169\]
\[a_{1,2} = \frac{- 3 \pm 13}{2} = - 8;5; \Longrightarrow\]
\[\left\{ \begin{matrix} x^{2} + x - 1 = - 8 \\ x^{2} + x - 1 = 5\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x² + x + 7 = 0\ \ (1) \\ x² + x - 6 = 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ D = 1 - 4 \cdot 7 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет.\]
\[(2)\ D = 1 + 24 = 25,\ \ \]
\[x_{1,2} = \frac{- 1 \pm 5}{2} = - 3;2.\]
\[Ответ:x = - 3;x = 2.\]
\[Пусть\ \ 2x^{2} + x - 1 = a,\ \ \ тогда\ \ \]
\[a \cdot (a - 3) + 2 = 0\]
\[a^{2} - 3a + 2 = 0\]
\[D = 9 - 4 \cdot 2 = 1\]
\[a_{1,2} = \frac{3 \pm 1}{2} = 1;2.\]
\[\left\{ \begin{matrix} 2x^{2} + x - 2 = 0\ \ (1) \\ 2x^{2} + x - 3 = 0\ \ (2) \\ \end{matrix} \right.\ \]
\[(1)\ D = 1 + 16 = 17,\ \ \]
\[x_{1,2} = \frac{- 1 \pm \sqrt{17}}{4};\]
\[(2)\ D = 1 + 24 = 25,\ \]
\[\ x_{3,4} = \frac{- 1 \pm 5}{4} = - 1,5;\ \ 1.\]
\[Ответ:x = 1;\ \ x = - 1,5;\ \ \]
\[x = \frac{- 1 \pm \sqrt{17}}{4}.\]
