ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 70

\[\boxed{\mathbf{70.}}\]

\[\left\{ \begin{matrix} x^{2} + x - 1{> x}^{2} - 2x \\ - 1 \leq x^{2} + x - 1 \leq 1 \\ - 1 \leq x^{2} - 2x \leq 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1.\ x^{2} + x - 1 > x^{2} - 2x\]

\[3x > 1\]

\[x > \frac{1}{3}.\]

\[2.\ - 1 \leq x^{2} + x - 1 \leq 1\ \]

\[\left\{ \begin{matrix} x^{2} + x - 1 \geq - 1 \\ x^{2} + x - 1 \leq 1\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} + x \geq 0\ \ \ \ \ \ \ \\ x^{2} + x - 2 \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} + x - 2 \leq 0\]

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

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

\[(x + 2)(x - 1) \leq 0\]

\[\left\{ \begin{matrix} x(x + 1) \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ (x + 2)(x - 1) \leq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x \leq - 1 \\ x \geq 0\ \ \ \\ x \geq - 2 \\ x \leq 1\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 2 \leq x < - 1 \\ 0 \leq x \leq 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3.\ - 1 \leq x^{2} - 2x \leq 1\]

\[\left\{ \begin{matrix} x^{2} - 2x \geq - 1 \\ x^{2} - 2x \leq 1\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 2x + 1 \geq 0 \\ x^{2} - 2x - 1 \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 2x - 1 \leq 0\]

\[D_{1} = 1 + 1 = 2\]

\[x_{1} = 1 + \sqrt{2};\]

\[x_{2} = 1 - \sqrt{2}.\]

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

\[\left\{ \begin{matrix} x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1 - \sqrt{2} \leq x \leq 1 + \sqrt{2} \\ \end{matrix} \right.\ \]

\[1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}.\]

\[Пересечение:\]

\[x \in \left( \frac{1}{3};1 \right\rbrack.\]

\[Ответ:x \in \left( \frac{1}{3};1 \right\rbrack.\]

\[\left\{ \begin{matrix} x^{2} - 4x + 3 < \frac{x - 1}{2}\text{\ \ \ \ } \\ - 1 \leq x^{2} - 4x + 3 \leq 1 \\ - 1 \leq \frac{x - 1}{2} \leq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1.\ x^{2} - 4x + 3 < \frac{x - 1}{2}\ \]

\[2x^{2} - 8x + 6 < x - 1\]

\[2x^{2} - 9x + 7 < 0\]

\[D = 81 - 56 = 25\]

\[x_{1} = \frac{9 + 5}{4} = 3,5;\]

\[x_{2} = \frac{9 - 5}{4} = 1.\]

\[(x - 1)(x - 3,5) < 0\]

\[1 < x < 3,5.\]

\[2.\ - 1 \leq x^{2} - 4x + 3 \leq 1\]

\[\left\{ \begin{matrix} x^{2} - 4x + 3 \geq - 1 \\ x^{2} - 4x + 3 \leq 1\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 4x + 4 \geq 0 \\ x^{2} - 4x + 2 \leq 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 4x + 2 \leq 0\]

\[D_{1} = 4 - 2 = 2\]

\[x_{1} = 2 + \sqrt{2};\]

\[x_{2} = 2 - \sqrt{2}.\]

\[\left\{ \begin{matrix} (x - 2)^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( x - 2 + \sqrt{2} \right)\left( x - 2 - \sqrt{2} \right) \leq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 - \sqrt{2} \leq x \leq 2 + \sqrt{2} \\ \end{matrix} \right.\ \]

\[2 - \sqrt{2} \leq x \leq 2 + \sqrt{2}.\]

\[3.\ - 1 \leq \frac{x - 1}{2} \leq 1\]

\[\left\{ \begin{matrix} \frac{x - 1}{2} \geq - 1 \\ \frac{x - 1}{2} \leq 1\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x - 1 \geq - 2 \\ x - 1 \leq 2\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x \geq - 1 \\ x \leq 3\ \ \ \ \\ \end{matrix} \right.\ \]

\[- 1 \leq x \leq 3.\]

\[Пересечение:\]

\[1 < x \leq 3.\]

\[Ответ:x \in (1;3\rbrack.\]

\[x \in R:\]

\[2x^{2} + 1 > 2x^{2} - x\]

\[- x < 1\]

\[x > - 1.\]

\[Ответ:x > - 1.\]

\[x \in R:\]

\[\frac{x - 2}{2} < \frac{4 - x}{2}\]

\[{x - 2 < 4 - x }{2x < 6}\]

\[x < 3.\]

\[Ответ:x < 3.\]