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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 153
Задание 153
\[\boxed{\text{153\ (153).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ |x - 3| + x = 15\]
\[|x - 3| = 15 - x\]
\[\left\{ \begin{matrix} x - 3 = 15 - x \\ 15 - x \geq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x = 18 \\ x \leq 15\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 9\ \ \ \\ x \leq 15 \\ \end{matrix} \right.\ \Longrightarrow \ \ x = 9\]
\[\left\{ \begin{matrix} - x + 3 = 15 - x \\ 15 - x \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 0 \cdot x = - 12 \\ x \leq 15\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \ \ нет\ корней.\]
\[Ответ:x = 9.\]
\[2)\ |x + 1| - 4x = 14\]
\[\left\{ \begin{matrix} x + 1 = 4x + 14 \\ 4x + 14 \geq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} - 3x = 13 \\ x \geq - 3,5\ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = - 4\frac{1}{3} \\ x \geq - 3,5 \\ \end{matrix} \right.\ \Longrightarrow \text{\ \ }\]
\[\Longrightarrow не\ удовлетворяет.\]
\[\left\{ \begin{matrix} - x - 1 = 4x + 14 \\ x \geq - 3,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} - 5x = 15 \\ x \geq - 3,5\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = - 3\ \ \ \ \\ x \geq - 3,5 \\ \end{matrix} \right.\ \Longrightarrow \ \ x = - 3\]
\[Ответ:\ x = - 3.\ \]
\[3)\ |3x - 12| - 2x = 1\]
\[\left\{ \begin{matrix} 3x - 12 = 1 + 2x \\ 1 + 2x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 13\ \ \ \ \\ x \geq - 0,5 \\ \end{matrix} \right.\ \Longrightarrow \ \ x = 13\]
\[\left\{ \begin{matrix} - 3x + 12 = 1 + 2x \\ x \geq - 0,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} - 5x = - 11 \\ x \geq - 0,5\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = \frac{11}{5}\text{\ \ \ \ } \\ x \geq - 0,5 \\ \end{matrix} \right.\ \Longrightarrow \ \ x = 2,2\]
\[Ответ:x = 13;x = 2,2.\ \]
\[4)\ |x + 2| - x = 1\]
\[\left\{ \begin{matrix} x + 2 = 1 + x \\ 1 + x \geq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 0 \cdot x = - 1 \\ x \geq - 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \ \ нет\ корней\]
\[\left\{ \begin{matrix} - x - 2 = 1 + x \\ x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} - 2x = 3 \\ x \geq - 1\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = - 1,5 \\ x \geq - 1\ \ \ \\ \end{matrix}\ \right.\ \Longrightarrow\]
\[\Longrightarrow не\ удовлетворяет.\]
\[Ответ:нет\ корней.\]
\[\boxed{\mathbf{153.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ x^{3} - 3x^{2} + 6x - 8 = 0\]
\[\left( x^{3} - 8 \right) - \left( 3x^{2} - 6x \right) = 0\]
\[(x - 2)\left( x^{2} + 2x + 4 \right) - 3x(x - 2) = 0\]
\[(x - 2)\left( x^{2} + 2x + 4 - 3x \right) = 0\]
\[(x - 2)\left( x^{2} - x + 4 \right) = 0\]
\[x^{2} - x + 4 = 0\]
\[D = 1 - 16 = - 15 < 0\]
\[нет\ корней.\]
\[x - 2 = 0\]
\[x = 2.\]
\[Ответ:\ \ x = 2.\]
\[2)\ x^{2} - 6x + 9 - x^{4} = 0\]
\[(x - 3)^{2} - \left( x^{2} \right)^{2} = 0\]
\[\left( x - 3 - x^{2} \right)\left( x - 3 + x^{2} \right) = 0\]
\[- x^{2} + x - 3 = 0\]
\[x^{2} - x + 3 = 0\]
\[D = 1 - 12 < 0\]
\[нет\ корней.\]
\[x^{2} + x - 3 = 0\]
\[D = 1 + 12 = 13\]
\[x = \frac{- 1 \pm \sqrt{13}}{2}.\]
\[Ответ:x = \frac{- 1 \pm \sqrt{13}}{2}.\]
\[3)\ x^{3} - 3x^{2} + 4 = 0\]
\[x^{3} - 3x^{2} + 1 + 3 = 0\]
\[\left( x^{3} + 1 \right) - 3\left( x^{2} - 1 \right) = 0\]
\[(x + 1)\left( x^{2} - x + 1 \right) - 3(x - 1)(x + 1) = 0\]
\[(x + 1)\left( x^{2} - x + 1 - 3x + 3 \right) = 0\]
\[(x + 1)\left( x^{2} - 4x + 4 \right) = 0\]
\[(x + 1)(x - 2)^{2} = 0\]
\[x = - 1;\ \ x = 2.\]
\[Ответ:x = - 1;\ \ x = 2.\]
\[4)\ x^{3} - 3x + 2 = 0\]
\[x^{3} - x - 2x + 2 = 0\]
\[x\left( x^{2} - 1 \right) - 2(x - 1) = 0\]
\[x(x - 1)(x + 1) - 2(x - 1) = 0\]
\[(x - 1)\left( x^{2} + x - 2 \right) = 0\]
\[x^{2} + x - 2 = 0\]
\[D = 1 + 8 = 9\]
\[x_{1} = \frac{( - 1 + 3)}{2} = 1;\]
\[x_{2} = \frac{- 1 - 3}{2} = - 2.\]
\[x - 1 = 0\]
\[x = 1.\]
\[Ответ:x = - 2;\ \ x = 1.\]
