Найдите значение дроби: (8y-56)/(y^2-27y+140) при x=-4; 22,5; 24.

Ответ:

\[\frac{8y - 56}{y^{2} - 27y + 140} =\]

\[= \frac{8 \cdot (y - 7)}{(y - 7)(y - 20)} = \frac{8}{y - 20}\]

\[y^{2} - 27y + 140 =\]

\[= (y - 7)(y - 20)\]

\[y_{1} + y_{2} = 27;\ \ \ y_{1} \cdot y_{2} = 140\]

\[y_{1} = 20;\ \ y_{2} = 7.\]

\[y = - 4:\]

\[\frac{8}{y - 20} = \frac{8}{- 24} = - \frac{1}{3}.\]

\[y = 22,5:\]

\[\frac{8}{y - 20} = \frac{8}{22,5 - 20} = \frac{8}{2,5} =\]

\[= \frac{80}{25} = \frac{360}{100} = 3,6.\]

\[y = 24:\]

\[\frac{8}{y - 20} = \frac{8}{24 - 20} = \frac{8}{4} = 2.\]