Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 7

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\[\textbf{а)}\ \sqrt{2x - 1} < x\]

\[2x - 1 \geq 0\]

\[2x \geq 1\]

\[x \geq 0,5.\]

\[M = \lbrack 0,5; + \infty).\]

\[2x - 1 < x^{2}\ \]

\[x^{2} - 2x + 1 > 0\]

\[(x - 1)^{2} > 0\]

\[x \neq 1.\]

\[x \in \lbrack 0,5;1) \cup (1; + \infty).\]

\[Ответ:\ x \in \lbrack 0,5;1) \cup (1; + \infty).\]

\[\textbf{б)}\ 2\sqrt{x - 1} < x\]

\[x - 1 \geq 0\]

\[x \geq 1.\]

\[M = \lbrack 1; + \infty).\]

\[4(x - 1) < x^{2}\]

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

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

\[(x - 2)^{2} > 0\]

\[x \neq 2.\]

\[x \in \lbrack 1;2) \cup (2; + \infty).\]

\[Ответ:\ x \in \lbrack 1;2) \cup (2; + \infty).\]

\[\textbf{в)}\ \sqrt{6x - 9} < x\]

\[6x - 9 \geq 0\]

\[6x \geq 9\]

\[x \geq 1,5.\]

\[M = \lbrack 1,5; + \infty).\]

\[6x - 9 < x^{2}\]

\[x^{2} - 6x + 9 > 0\]

\[(x - 3)^{2} > 0\]

\[x \neq 3.\]

\[x \in \lbrack 1,5;3) \cup (3; + \infty).\]

\[Ответ:\ x \in \lbrack 1,5;3) \cup (3; + \infty)\]

\[\textbf{г)}\ 2\sqrt{2x - 4} < x\]

\[2x - 4 \geq 0\]

\[2x \geq 4\]

\[x \geq 2.\]

\[M = \lbrack 2; + \infty).\]

\[4(2x - 4) < x^{2}\]

\[8x - 16 < x^{2}\]

\[x^{2} - 8x + 16 > 0\]

\[(x - 4)^{2} > 0\]

\[x \neq 4.\]

\[x \in \lbrack 2;4) \cup (4; + \infty).\]

\[Ответ:\ x \in \lbrack 2;4) \cup (4; + \infty)\]