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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