Решебник по алгебре 9 класс Макарычев Задание 221

 

\[\boxed{\text{221\ (221).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)} - x^{2} + 20x - 103 =\]

\[= - \left( x^{2} - 20x + 103 \right) =\]

\[= - \left( x^{2} - 2 \cdot 10 \cdot x + 100 + 3 \right) =\]

\[= - \left( (x - 10)^{2} + 3 \right) < 0;\ \ \ \ \ \]

\[так\ как:\ \]

\[(x - 10)^{2} + 3 > 0.\]

\[\textbf{б)}\ x^{2} - 16x + 65 =\]

\[= x^{2} - 2 \cdot 8 \cdot x + 64 + 1 =\]

\[= (x - 8)^{2} + 1 > 0.\]

\[\boxed{\text{221.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[Пусть\ 2x^{2} + 3 = t,\ \ тогда\ \]

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

\[D = 36 - 11 = 25\]

\[t_{1,2} = 6 \pm 5 = 11;1:\]

\[\left\{ \begin{matrix} 2x^{2} + 3 = 11 \\ 2x^{2} + 3 = 1\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 2x^{2} = 8 \\ 2x^{2} = - 1 \\ \end{matrix}\ \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x² = 4\ \ \ \\ x² = - 1 \\ \end{matrix} \right.\ \]

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

\[x² = - 1 \Longrightarrow корней\ не\ имеет.\]

\[Ответ:x = \pm 2.\]

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

\[Пусть\ \ t^{2} - 2t = y,\ \ тогда\ \ \]

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

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

\[D = 1 + 3 = 4\]

\[y_{1,2} = 1 \pm 2 = 3;\ - 1:\]

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

\[\Longrightarrow \left\{ \begin{matrix} t² - 2t - 3 = 0\ \ (1) \\ t² - 2t + 1 = 0\ \ (2) \\ \end{matrix} \right.\ \]

\[(1)\text{\ \ }D = 1 + 3 = 4,\ \ \]

\[t_{1,2} = 1 \pm 2 = 3;\ - 1;\]

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

\[t_{3} = 1.\]

\[Ответ:t = 3;t = \pm 1.\]

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

\[= 40\]

\[Пусть\ x^{2} + x - 1 = a,\ \ тогда\ \]

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

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

\[D = 9 + 160 = 169\]

\[a_{1,2} = \frac{- 3 \pm 13}{2} = - 8;5; \Longrightarrow\]

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

\[\Longrightarrow \left\{ \begin{matrix} x² + x + 7 = 0\ \ (1) \\ x² + x - 6 = 0\ \ (2) \\ \end{matrix} \right.\ \]

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

\[\Longrightarrow корней\ нет.\]

\[(2)\ D = 1 + 24 = 25,\ \ \]

\[x_{1,2} = \frac{- 1 \pm 5}{2} = - 3;2.\]

\[Ответ:x = - 3;x = 2.\]

\[Пусть\ \ 2x^{2} + x - 1 = a,\ \ \ тогда\ \ \]

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

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

\[D = 9 - 4 \cdot 2 = 1\]

\[a_{1,2} = \frac{3 \pm 1}{2} = 1;2.\]

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

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

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

\[(2)\ D = 1 + 24 = 25,\ \]

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

\[Ответ:x = 1;\ \ x = - 1,5;\ \ \]

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