ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 929

\[1)\ \frac{\sqrt{3x^{3} - 22x^{2} + 40x}}{x - 4} \geq 3x - 10\]

\[x > 4:\]

\[\sqrt{3x^{3} - 22x^{2} + 40x} \geq (3x - 10)(x - 4)\]

\[\sqrt{x\left( 3x^{2} - 22x + 40 \right)} \geq 3x^{2} - 22x + 40\]

\[\sqrt{x} \geq \sqrt{3x^{2} - 22x + 40}\]

\[x \geq 3x^{2} - 22x + 40\]

\[3x^{2} - 23x + 40 \leq 0\]

\[D = 529 - 480 = 49\]

\[x_{1} = \frac{23 - 7}{2 \bullet 3} = \frac{8}{3};\]

\[x_{2} = \frac{23 + 7}{2 \bullet 3} = 5;\]

\[\left( x - \frac{8}{3} \right)(x - 5) \leq 0\]

\[\frac{8}{3} \leq x \leq 5.\]

\[x < 4:\]

\[\sqrt{3x^{3} - 22x^{2} + 40x} \leq (3x - 10)(x - 4)\]

\[x \leq \frac{8}{3};\ \ \ x \geq 5.\]

\[Еще\ корни:\]

\[3x^{2} - 22x + 40 = 0\]

\[D = 484 - 480 = 4\]

\[x_{1} = \frac{22 - 2}{2 \bullet 3} = \frac{10}{3};\ \]

\[x_{2} = \frac{22 + 2}{2 \bullet 3} = 4.\]

\[Область\ определения:\]

\[3x^{3} - 22x^{2} + 40x \geq 0\]

\[x\left( x - \frac{10}{3} \right)(x - 4) \geq 0\]

\[0 \leq x \leq \frac{10}{3}.\]

\[x - 4 \neq 0\]

\[x \neq 4\]

\[x > 4.\]

\[Ответ:\ \ \]

\[x \in \left\lbrack 0;\ \frac{8}{3} \right\rbrack \cup \left\{ \frac{10}{3} \right\} \cup (4;\ 5\rbrack.\]

\[2)\ \frac{\sqrt{2x^{3} - 22x^{2} + 60x}}{x - 6} \geq 2x - 10\]

\[x > 6:\]

\[\sqrt{2x^{3} - 22x^{2} + 60x} \geq (2x - 10)(x - 6)\]

\[\sqrt{x\left( 2x^{2} - 22x + 60 \right)} \geq 2x^{2} - 22x + 60\]

\[\sqrt{x} \geq \sqrt{2x^{2} - 22x + 60}\]

\[x \geq 2x^{2} - 22x + 60\]

\[2x^{2} - 23x + 60 \leq 0\]

\[D = 529 - 480 = 49\]

\[x_{1} = \frac{23 - 7}{2 \bullet 2} = 4;\]

\[x_{2} = \frac{23 + 7}{2 \bullet 2} = 7,5;\]

\[(x - 4)(x - 7,5) \leq 0\]

\[4 \leq x \leq 7,5.\]

\[x < 6:\]

\[\sqrt{2x^{3} - 22x^{2} + 60x} \leq (2x - 10)(x - 6)\]

\[x \leq 4;\ \ \ x \geq 7,5.\]

\[Еще\ корни:\]

\[2x^{2} - 22x + 60 = 0\]

\[x^{2} - 11x + 30 = 0\]

\[D = 121 - 120 = 1\]

\[x_{1} = \frac{11 - 1}{2} = 5;\]

\[x_{2} = \frac{11 + 1}{2} = 6.\]

\[Область\ определения:\]

\[2x^{3} - 22x^{2} + 60x \geq 0\]

\[x(x - 5)(x - 6) \geq 0\]

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

\[x - 6 \neq 0\]

\[x \neq 6\]

\[x > 6.\]

\[Ответ:\ \ \]

\[x \in \lbrack 0;\ 4\rbrack \cup \left\{ 5 \right\} \cup (6;\ 7,5\rbrack.\]