Вопрос:

Решить неравенства методом интервалов: (5-3x)^2*(x^2-10x+9)<=0.

Ответ:

\[(5 - 3x)^{2}\left( x^{2} - 10x + 9 \right) \leq 0\]

\[x^{2} - 10x + 9 = (x - 1)(x - 9)\]

\[D_{1} = 25 - 9 = 16\]

\[x_{1} = 5 + 4 = 9;\]

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

\[(3x - 5)^{2}(x - 1)(x - 9) \leq 0\]

\[3\left( x - \frac{5}{3} \right)^{2}(x - 1)(x - 9) \leq 0\]

\[1 \leq x \leq 9.\]

\[Ответ:1 \leq x \leq 9.\]

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