a: ĐKXĐ: x∉{0;-6;6;3}
b: \(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\)
\(=\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\)
\(=\frac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}=\frac{12x-36}{x\left(x-6\right)\left(x+6\right)}=\frac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\)
Ta có: \(A=\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
\(=\frac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\cdot\frac{x\left(x+6\right)}{2\left(x-3\right)}-\frac{x}{x-6}=\frac{6}{x-6}-\frac{x}{x-6}=-1\)
d: Vì A=-1 là số nguyên
với mọi x thỏa mãn ĐKXĐ, A luôn là số nguyên
=>x∈Z\{0;6;-6;3}
