Question

15b-4/(b^2-9)(b+1) найдите допустимые значения переменной​

Answer 1

Ответ:

$\dfrac{15b-4}{(b^2-9)(b+1)}\ \ \ \Rightarrow \ \ \ \left\{\begin{array}{l}b^2-9\ne 0\\b+1\ne 0\end{array}\right\ \ \left\{\begin{array}{l}(b-3)(b+3)\ne 0\\b\ne -1\end{array}\right\ \ \left\{\begin{array}{l}b\ne -3\ ,\ b\ne 3\\b\ne -1\end{array}\right$

$\boxed{\ b\in (-\infty ;-3)\cup (-3;-1)\cup (-1;3)\cup (3;+\infty )\ }$