26. Найдите значение дроби: a) 36-x² / 6-7x+x² при х = -9; -99; -999; б) 4x² + 8x - 32 / 4x²-16 при х = -1; 5; 10.

a)

\[\frac{36-x^2}{6-7x+x^2}\]

При \(x = -9\):

\[\frac{36-(-9)^2}{6-7(-9)+(-9)^2} = \frac{36-81}{6+63+81} = \frac{-45}{150} = -\frac{3}{10} = -0.3\]

При \(x = -99\):

\[\frac{36-(-99)^2}{6-7(-99)+(-99)^2} = \frac{36-9801}{6+693+9801} = \frac{-9765}{10500} = -\frac{651}{700} = -0.93\]

При \(x = -999\):

\[\frac{36-(-999)^2}{6-7(-999)+(-999)^2} = \frac{36-998001}{6+6993+998001} = \frac{-997965}{1005000} = -\frac{199593}{201000} = -0.993\]

б)

\[\frac{4x^2 + 8x - 32}{4x^2-16}\]

При \(x = -1\):

\[\frac{4(-1)^2 + 8(-1) - 32}{4(-1)^2-16} = \frac{4 - 8 - 32}{4-16} = \frac{-36}{-12} = 3\]

При \(x = 5\):

\[\frac{4(5)^2 + 8(5) - 32}{4(5)^2-16} = \frac{100 + 40 - 32}{100-16} = \frac{108}{84} = \frac{9}{7} = 1\frac{2}{7} \approx 1.29\]

При \(x = 10\):

\[\frac{4(10)^2 + 8(10) - 32}{4(10)^2-16} = \frac{400 + 80 - 32}{400-16} = \frac{448}{384} = \frac{7}{6} = 1\frac{1}{6} \approx 1.17\]

Ответ: a) -0.3; -0.93; -0.993 б) 3; 1.29; 1.17