Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 6

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\[\textbf{а)}\ \sqrt{3x - 2} < x\]

\[3x - 2 \geq 0\]

\[3x \geq 2\]

\[x \geq \frac{2}{3}.\]

\[M = \left\lbrack \frac{2}{3}; + \infty \right).\]

\[3x - 2 < x^{2}\ \]

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

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

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

\[(x - 1)(x - 2) > 0\]

\[x < 1;\ \ \ x > 2.\]

\[x \in \left\lbrack \frac{2}{3};1 \right) \cup (2; + \infty).\]

\[Ответ:\ x \in \left\lbrack \frac{2}{3};1 \right) \cup (2; + \infty).\]

\[\textbf{б)}\ \sqrt{4x - 3} < x\]

\[4x - 3 \geq 0\]

\[4x \geq 3\]

\[x \geq \frac{3}{4}.\]

\[M = \left\lbrack \frac{3}{4}; + \infty \right).\]

\[4x - 3 < x^{2}\]

\[x^{2} - 4x + 3 > 0\]

\[D_{1} = 4 - 3 = 1\]

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

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

\[(x - 1)(x - 3) > 0\]

\[x < 1;\ \ x > 3.\]

\[x \in \left\lbrack \frac{3}{4};1 \right) \cup (3; + \infty).\]

\[Ответ:\ x \in \left\lbrack \frac{3}{4};1 \right) \cup (3; + \infty).\]

\[\textbf{в)}\ \sqrt{5x - 4} < x\]

\[5x - 4 \geq 0\]

\[5x \geq 4\]

\[x \geq 0,8.\]

\[M = \lbrack 0,8;\ + \infty).\]

\[5x - 4 < x^{2}\]

\[x^{2} - 5x + 4 > 0\]

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 4\]

\[x_{1} = 1;\ \ x_{2} = 4.\]

\[(x - 1)(x - 4) > 0\]

\[x < 1;\ \ \ \ x > 4.\]

\[x \in \lbrack 0,8;1) \cup (4; + \infty).\]

\[\textbf{г)}\ \sqrt{6x - 5} < x\]

\[6x - 5 \geq 0\]

\[6x \geq 5\]

\[x \geq \frac{5}{6}.\]

\[M = \left\lbrack \frac{5}{6}; + \infty \right).\]

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

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

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

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

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

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

\[x < 1;\ \ \ x > 5.\]

\[x \in \left\lbrack \frac{5}{6};1 \right) \cup (5; + \infty).\]