Вопрос:

Найдите значение дроби: (x^2-18x+80)/(5x-50) при x=-12; 8,5; 48.

Ответ:


\[\frac{x^{2} - 18x + 80}{5x - 50} =\]


\[= \frac{(x - 10)(x - 8)}{5(x - 10)} = \frac{x - 8}{5}\ \]


\[x^{2} - 18x + 80 = 0\]


\[x_{1} + x_{2} = 18;\ \ \ \ x_{1} \cdot x_{2} = 80\]


\[x_{1} = 10;\ \ \ x_{2} = 8.\]


\[x = - 12:\]


\[\frac{x - 8}{5} = \frac{- 12 - 8}{5} = - \frac{20}{5} = - 4.\]


\[x = 8,5:\]


\[\frac{x - 8}{5} = \frac{8,5 - 8}{5} = \frac{0,5}{5} = \frac{5}{50} =\]


\[= 0,1.\]


\[x = 48:\]


\[\frac{x - 8}{5} = \frac{48 - 8}{5} = \frac{40}{5} = 8.\]


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