Вопрос:

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

Ответ:


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


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


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


\[D = 16 + 33 = 49\]


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


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


\[x = - 9:\]


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


\[x = 12:\]


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


\[x = 111:\]


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


Подать жалобу Правообладателю

Похожие