a: \(=\dfrac{x^2+5x-36-x^2-4x}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{\left(x-4\right)\left(x+4\right)}{x-5}\)
\(=\dfrac{x-36}{x-5}\)
b: |x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=4(loại) hoặc x=-2(nhận)
Khi x=-2 thì \(A=\dfrac{-2-36}{-2-5}=\dfrac{-38}{-7}=\dfrac{38}{7}\)
c: Đê A nguyên thì x-5-31 chia hết cho x-5
=>\(x-5\in\left\{1;-1;31;-31\right\}\)
=>\(x\in\left\{6;36;-26\right\}\)