Вопрос:

Найдите значение дроби: (x^2-8x-33)/(10x+30) при x=-9; 12; 111.

Ответ:

\[\frac{x^{2} - 8x - 33}{10x + 30} =\]

\[= \frac{(x + 3)(x - 11)}{10 \cdot (x + 3)} = \frac{x - 11}{10}\]

\[x^{2} - 8x - 33 = (x + 3)(x - 11)\]

\[D = 16 + 33 = 49\]

\[x_{1} = 4 + 7 = 11;\ \ \ \]

\[x_{2} = 4 - 7 = - 3.\]

\[x = - 9:\]

\[\frac{x - 11}{10} = \frac{- 9 - 11}{10} = - 2.\]

\[x = 12:\]

\[\frac{x - 11}{10} = \frac{12 - 11}{10} = \frac{1}{10} = 0,1.\]

\[x = 111:\]

\[\frac{x - 11}{10} = \frac{111 - 11}{10} = 10.\]

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