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МатематикаРешебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 70
Задание 70
\[\boxed{\mathbf{70.}}\]
\[\left\{ \begin{matrix} x^{2} + x - 1{> x}^{2} - 2x \\ - 1 \leq x^{2} + x - 1 \leq 1 \\ - 1 \leq x^{2} - 2x \leq 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1.\ x^{2} + x - 1 > x^{2} - 2x\]
\[3x > 1\]
\[x > \frac{1}{3}.\]
\[2.\ - 1 \leq x^{2} + x - 1 \leq 1\ \]
\[\left\{ \begin{matrix} x^{2} + x - 1 \geq - 1 \\ x^{2} + x - 1 \leq 1\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + x \geq 0\ \ \ \ \ \ \ \\ x^{2} + x - 2 \leq 0 \\ \end{matrix} \right.\ \]
\[x^{2} + x - 2 \leq 0\]
\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = - 2;\ \ x_{2} = 1.\]
\[(x + 2)(x - 1) \leq 0\]
\[\left\{ \begin{matrix} x(x + 1) \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ (x + 2)(x - 1) \leq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \leq - 1 \\ x \geq 0\ \ \ \\ x \geq - 2 \\ x \leq 1\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} - 2 \leq x < - 1 \\ 0 \leq x \leq 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[3.\ - 1 \leq x^{2} - 2x \leq 1\]
\[\left\{ \begin{matrix} x^{2} - 2x \geq - 1 \\ x^{2} - 2x \leq 1\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 2x + 1 \geq 0 \\ x^{2} - 2x - 1 \leq 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 2x - 1 \leq 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1} = 1 + \sqrt{2};\]
\[x_{2} = 1 - \sqrt{2}.\]
\[\left\{ \begin{matrix} (x - 1)^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( x - 1 - \sqrt{2} \right)\left( x - 1 + \sqrt{2} \right) \leq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1 - \sqrt{2} \leq x \leq 1 + \sqrt{2} \\ \end{matrix} \right.\ \]
\[1 - \sqrt{2} \leq x \leq 1 + \sqrt{2}.\]
\[Пересечение:\]
\[x \in \left( \frac{1}{3};1 \right\rbrack.\]
\[Ответ:x \in \left( \frac{1}{3};1 \right\rbrack.\]
\[\left\{ \begin{matrix} x^{2} - 4x + 3 < \frac{x - 1}{2}\text{\ \ \ \ } \\ - 1 \leq x^{2} - 4x + 3 \leq 1 \\ - 1 \leq \frac{x - 1}{2} \leq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1.\ x^{2} - 4x + 3 < \frac{x - 1}{2}\ \]
\[2x^{2} - 8x + 6 < x - 1\]
\[2x^{2} - 9x + 7 < 0\]
\[D = 81 - 56 = 25\]
\[x_{1} = \frac{9 + 5}{4} = 3,5;\]
\[x_{2} = \frac{9 - 5}{4} = 1.\]
\[(x - 1)(x - 3,5) < 0\]
\[1 < x < 3,5.\]
\[2.\ - 1 \leq x^{2} - 4x + 3 \leq 1\]
\[\left\{ \begin{matrix} x^{2} - 4x + 3 \geq - 1 \\ x^{2} - 4x + 3 \leq 1\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 4x + 4 \geq 0 \\ x^{2} - 4x + 2 \leq 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 4x + 2 \leq 0\]
\[D_{1} = 4 - 2 = 2\]
\[x_{1} = 2 + \sqrt{2};\]
\[x_{2} = 2 - \sqrt{2}.\]
\[\left\{ \begin{matrix} (x - 2)^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left( x - 2 + \sqrt{2} \right)\left( x - 2 - \sqrt{2} \right) \leq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 - \sqrt{2} \leq x \leq 2 + \sqrt{2} \\ \end{matrix} \right.\ \]
\[2 - \sqrt{2} \leq x \leq 2 + \sqrt{2}.\]
\[3.\ - 1 \leq \frac{x - 1}{2} \leq 1\]
\[\left\{ \begin{matrix} \frac{x - 1}{2} \geq - 1 \\ \frac{x - 1}{2} \leq 1\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - 1 \geq - 2 \\ x - 1 \leq 2\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq - 1 \\ x \leq 3\ \ \ \ \\ \end{matrix} \right.\ \]
\[- 1 \leq x \leq 3.\]
\[Пересечение:\]
\[1 < x \leq 3.\]
\[Ответ:x \in (1;3\rbrack.\]
\[x \in R:\]
\[2x^{2} + 1 > 2x^{2} - x\]
\[- x < 1\]
\[x > - 1.\]
\[Ответ:x > - 1.\]
\[x \in R:\]
\[\frac{x - 2}{2} < \frac{4 - x}{2}\]
\[{x - 2 < 4 - x }{2x < 6}\]
\[x < 3.\]
\[Ответ:x < 3.\]
