Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1578

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

\[x > 4:\]

\[2\frac{2}{3} \leq x \leq 3\frac{1}{3};\text{\ \ }4 \leq x \leq 5.\]

\[x < 4:\]

\[\sqrt{x\left( 3x^{2} - 12x - 10x + 40 \right)} \leq\]

\[\leq (3x - 10)(x - 4)\]

\[\left( x - \frac{8}{3} \right)(3x - 10)(x - 4)(x - 5) \geq 0\]

\[x \leq 2\frac{2}{3};\ 3\frac{1}{3} \leq x \leq 4;x \geq 5.\]

\[Имеет\ смысл\ при:\]

\[x - 4 \neq 0\]

\[x \neq .\]

\[x\left( 3x^{2} - 22x + 40 \right) \geq 0\]

\[x(3x - 10)(x - 4) \geq 0\]

\[0 \leq x \leq 3\frac{1}{3};\text{\ \ }\]

\[x > 4.\]

\[Ответ:\ \ 0 \leq x \leq 2\frac{2}{3};\ \ \]

\[x = 3\frac{1}{3};\ \ 4 < x \leq 5.\]