a: ĐKXĐ: x∉{0;2;-2}
\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x^2}{x\left(x^2-4\right)}-\frac{6}{3x-6}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}=\frac{2x-2-2x-4}{6\left(x-2\right)}=\frac{-1}{x-2}\)
b: \(\left|x\right|=\frac12\)
=>\(\left[\begin{array}{l}x=\frac12\\ x=-\frac12\end{array}\right.\)
Thay x=1/2 vào A, ta được: \(A=-\frac{1}{\frac12-2}=-1:\frac{-3}{2}=\frac23\)
THay x=-1/2 vào A, ta được: \(A=\frac{-1}{-\frac12-2}=-1:\frac{-5}{2}=\frac25\)
c: Để A là số nguyên thì -1⋮x-2
=>x-2∈{1;-1}
=>x∈{3;1}
