Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 45

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\[\textbf{а)}\ \sqrt{2x - 1} > x - 2\]

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

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

\[x^{2} - 6x + 5 < 0\]

\[D_{1} = 9 - 5 = 4\]

\[x_{1} = 3 + 2 = 5;\]

\[x_{2} = 3 - 2 = 1;\]

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

\[1 < x < 5.\]

\[\left\{ \begin{matrix} x > 1 \\ x < 5 \\ x \geq 2 \\ \end{matrix} \right.\ \]

\[2 \leq x < 5.\]

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

\[\left\{ \begin{matrix} x \geq 0,5 \\ x < 2\ \ \ \\ \end{matrix} \right.\ \]

\[0,5 \leq x < 2.\]

\[Решение\ неравенства:\]

\[0,5 \leq x < 5.\]

\[Ответ:0,5 \leq x < 5.\]

\[\textbf{б)}\ \sqrt{2x + 1} > x - 1\]

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

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

\[x^{2} - 4x < 0\]

\[x(x - 4) < 0\]

\[0 < x < 4.\]

\[\left\{ \begin{matrix} x > 0 \\ x \geq 1 \\ x < 4 \\ \end{matrix} \right.\ \]

\[1 \leq x < 4.\]

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

\[\left\{ \begin{matrix} x \geq - 0,5 \\ x < 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[- 0,5 \leq x < 1.\]

\[Решение\ неравенства:\]

\[- 0,5 \leq x < 4.\]

\[Ответ:\ - 0,5 \leq x < 4.\]