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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 28
Задание 28
\[\boxed{\mathbf{28.}}\]
\[\textbf{а)}\ \left| x^{2} - 4x + 2 \right| = - x^{2} + 6x - 6\]
\[1)\ \left\{ \begin{matrix} x^{2} - 4x + 2 = - x^{2} + 6x - 6 \\ - x^{2} + 6x - 6 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x^{2} - 10x + 8 = 0 \\ x^{2} - 6x + 6 \leq 0\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[2x^{2} - 10x + 8 = 0\ \ |\ :2\]
\[x^{2} - 5x + 4 = 0\]
\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 4\]
\[x_{1} = 4;\ \ x_{2} = 1.\]
\[x^{2} - 6x + 6 \leq 0\]
\[D_{1} = 9 - 6 = 3\]
\[x_{1} = 3 + \sqrt{3};\]
\[x_{2} = 3 - \sqrt{3}.\]
\[\left( x - 3 + \sqrt{3} \right)\left( x + 3 - \sqrt{3} \right) \leq 0\]
\[3 - \sqrt{3} < x < 3 + \sqrt{3}.\]
\[\left\{ \begin{matrix} x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 - \sqrt{3} < x < 3 + \sqrt{3} \\ \end{matrix} \right.\ \]
\[x = 4.\]
\[2)\ \left\{ \begin{matrix} x^{2} - 4x + 2 = - ( - x^{2} + 6x - 6) \\ - x^{2} + 6x - 6 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x - 4 = 0\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 6x + 6 \leq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 - \sqrt{3} < x < 3 + \sqrt{3} \\ \end{matrix} \right.\ \]
\[x = 2.\]
\[Ответ:x = 2;x = 4.\]
\[\textbf{б)}\ \left| x^{2} - 2x - 1 \right| = - x^{2} + 4x - 1\]
\[1)\ \left\{ \begin{matrix} x^{2} - 2x - 1 = - x^{2} + 4x - 1 \\ - x^{2} + 4x - 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x^{2} - 6x = 0\ \ \ \ \ \ \\ x^{2} - 4x + 1 \leq 0 \\ \end{matrix} \right.\ \]
\[2x^{2} - 6x = 0\]
\[x(x - 3) = 0\]
\[x = 0;\ \ x = 3.\]
\[x^{2} - 4x + 1 \leq 0\]
\[D_{1} = 4 - 1 = 3\]
\[x_{1} = 2 + \sqrt{3};\]
\[x_{2} = 2 - \sqrt{3}.\]
\[\left( x - 2 + \sqrt{3} \right)\left( x - 2 - \sqrt{3} \right) \leq 0\]
\[2 - \sqrt{3} < x < 2 + \sqrt{3}.\]
\[\left\{ \begin{matrix} x = 0;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 3 \\ 2 - \sqrt{3} < x < 2 + \sqrt{3} \\ \end{matrix} \right.\ \]
\[x = 3.\]
\[2)\ \left\{ \begin{matrix} x^{2} - 2x - 1 = - \left( - x^{2} \right) + 4x - 1 \\ - x^{2} + 4x - 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x - 2 = 0\ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4x + 1 \leq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 - \sqrt{3} < x < 2 + \sqrt{3} \\ \end{matrix} \right.\ \]
\[x = 1.\]
\[Ответ:x = 1;x = 3.\]
\[\textbf{в)}\ \left| x^{2} - 2^{x} - 8 \right| =\]
\[= x^{2} + 2^{x} - 10\]
\[1)\ \left\{ \begin{matrix} x^{2} - 2^{x} - 8 = x^{2} + 2^{x} - 10 \\ x^{2} + 2^{x} - 10 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2 \cdot 2^{x} = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2^{x} - 10 \geq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2^{x} = 2^{0}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} + 2^{x} - 10 \geq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2^{x} - 10 \geq 0 \\ \end{matrix} \right.\ \]
\[решений\ нет.\]
\[2)\ \left\{ \begin{matrix} x^{2} - 2^{x} - 8 = - \left( - x^{2} + 2^{x} - 10 \right) \\ x^{2} + 2^{x} - 10 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x^{2} = 18\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2^{x} - 10 \geq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 2^{x} - 10 \geq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 3;\ \ \ \ \ \ \ x = - 3 \\ x^{2} + 2^{x} - 10 \geq 0 \\ \end{matrix} \right.\ \]
\[x = 3.\]
\[Ответ:x = 3.\]
\[\textbf{г)}\ \left| x^{2} - 3^{x} - 1 \right| = x^{2} + 3^{x} - 7\]
\[1)\ \left\{ \begin{matrix} x^{2} - 3^{x} - 1 = x^{2} + 3^{x} - 7 \\ x^{2} + 3^{x} - 7 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2 \cdot 3^{x} = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3^{x} - 7 \geq 0\ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 3^{x} = 3^{1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} + 3^{x} - 7 \geq 0\ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3^{x} - 7 \geq 0\ \\ \end{matrix} \right.\ \]
\[нет\ решений.\]
\[2)\ \left\{ \begin{matrix} x^{2} - 3^{x} - 1 = - (x^{2} + 3^{x} - 7) \\ x^{2} + 3^{x} - 7 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 2x^{2} = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3^{x} - 7 \geq 0\ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3^{x} - 7 \geq 0\ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = - 2;\ \ \ \ \ x = 2 \\ x^{2} + 3^{x} - 7 \geq 0\ \\ \end{matrix} \right.\ \]
\[x = 2.\]
\[Ответ:x = 2.\]