ГДЗ по алгебре 9 класс Макарычев Задание 276

\[\boxed{\text{276}\text{\ (276)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( 2x^{2} + 3 \right)^{2} - 12 \cdot\]

\[\cdot \left( 2x^{2} + 3 \right) + 11 = 0\]

\[Пусть\ 2x^{2} + 3 = t:\ \]

\[t^{2} - 12t + 11 = 0\]

\[D_{1} = 36 - 11 = 25\]

\[t_{1} = 6 + 5 = 11;\ \]

\[\text{\ \ }t_{2} = 6 - 5 = 1\]

\[\left\{ \begin{matrix} 2x^{2} + 3 = 11 \\ 2x^{2} + 3 = 1\ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2x^{2} = 8\ \ \ \\ 2x^{2} = - 1 \\ \end{matrix}\ \right.\ \Longrightarrow\]

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

\[x^{2} = 4 \Longrightarrow x_{1,2} = \pm 2,\]

\[x^{2} = - 1 \Longrightarrow корней\ не\ имеет.\]

\[Ответ:x = \pm 2.\]

\[\textbf{б)}\ \left( t^{2} - 2t \right)^{2} - 3 = 2 \cdot (t^{2} - 2t)\]

\[Пусть\ \ t^{2} - 2t = y:\]

\[y^{2} - 3 = 2y\]

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

\[D_{1} = 1 + 3 = 4\]

\[y_{1} = 1 + 2 = 3;\ \ \ \ \ \]

\[\ y_{2} = 1 - 2 = - 1;\]

\[\left\{ \begin{matrix} t^{2} - 2t = 3\ \ \ \\ t^{2} - 2t = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} t^{2} - 2t - 3 = 0\ \ (1) \\ t^{2} - 2t + 1 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ t^{2} - 2t - 3 = 0\ \]

\[D_{1} = 1 + 3 = 4,\ \ \]

\[t_{1} = 1 + 2 = 3;\ \ \ t_{2} =\]

\[= 1 - 2 = - 1;\]

\[(2)\ t^{2} - 2t + 1 = 0\]

\[(t - 1)^{2} = 0\ \ \]

\[t - 1 = 0\]

\[t_{3} = 1.\]

\[Ответ:t = 3;t = \pm 1.\]

\[\textbf{в)}\ \left( x^{2} + x - 1 \right)\left( x^{2} + x + 2 \right) = 40\]

\[Пусть\ x^{2} + x - 1 = a:\text{\ \ }\]

\[a \cdot (a + 3) = 40\]

\[a^{2} + 3a - 40 = 0\]

\[D = 9 + 160 = 169\]

\[a_{1} = \frac{- 3 - 13}{2} = - 8;\ \ a_{2} =\]

\[= \frac{- 3 + 13}{2} = 5;\]

\[\left\{ \begin{matrix} x^{2} + x - 1 = - 8 \\ x^{2} + x - 1 = 5\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x^{2} + x + 7 = 0\ \ (1) \\ x^{2} + x - 6 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ x^{2} + x + 7 = 0\]

\[D = 1 - 4 \cdot 7 = - 27 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

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

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 6\]

\[x_{1} = - 3;\ \ \ \ x_{2} = 2.\]

\[Ответ:x = - 3;x = 2.\]

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

\[+ 2 = 0\]

\[Пусть\ \ 2x^{2} + x - 1 = a:\ \]

\[a \cdot (a - 3) + 2 = 0\]

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

\[a_{1} + a_{2} = 3;\ \ \ \ a_{1} \cdot a_{2} = 2\]

\[a_{1} = 1;\ \ \ a_{2} = 2.\]

\[\left\{ \begin{matrix} 2x^{2} + x - 2 = 0\ \ (1) \\ 2x^{2} + x - 3 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\ 2x^{2} + x - 2 = 0\]

\[D = 1 + 16 = 17,\ \ \]

\[x_{1,2} = \frac{- 1 \pm \sqrt{17}}{4}.\]

\[(2)\ 2x^{2} + x - 3 = 0\]

\[D = 1 + 24 = 25,\ \]

\[x_{3} = \frac{- 1 - 5}{4} = - 1,5;\ \ x_{4} =\]

\[= \frac{- 1 + 5}{4} = 1.\]

\[Ответ:x = 1;\ \ x = - 1,5;\ \ x =\]

\[= \frac{- 1 \pm \sqrt{17}}{4}.\]