\(ĐK:x\ne0;x\ne\pm2\\ a,A=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right]:\dfrac{x^2-4+10-x^2}{x+2}\\ A=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6\left(x-2\right)}=\dfrac{1}{2-x}\\ b,\left|x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\\x=-\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\end{matrix}\right.\)
\(c,A\in Z\Leftrightarrow2-x\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{1;3\right\}\left(tm\right)\)
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}
