Вопрос:

Решите систему неравенств (x^2-5x+6)^2<=0; (x^2+3x-1)^2<=100.

Ответ:

\[1)\ x^{2} - 5x + 6 = (x - 2)(x - 3)\]

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 6\]

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

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

\[x = 2;\ \ x = 3.\]

\[2)\ x = 2:\]

\[x^{2} + 3x - 1 = 4 + 6 - 1 = 9;\]

\[x = 3:\]

\[x^{2} + 3x - 1 = 9 + 9 - 1 =\]

\[= 17 > 10\ (не\ подходит)\text{.\ }\]

\[\left\{ \begin{matrix} x = 2;\ \ x = 3 \\ x = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:x = 2.\]

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