a) ĐKXĐ: \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{2+x}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(\Leftrightarrow A=\left[\frac{x}{x^2-4}+\frac{2\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}+\frac{2-x}{\left(2+x\right)\left(2-x\right)}\right]:\text{ }\left[\left(x-2\right)\left(x+2\right)+\frac{10-x^2}{x+2}\right]\)\(\Leftrightarrow A=\left(\frac{x-2x-4-2+x}{x^2-4}\right):\text{ }\left(\frac{10-x^2+x^2-4}{x+2}\right)\)
\(\Leftrightarrow A=\frac{-6}{\left(x+2\right)\left(x-2\right)}:\frac{6}{x+2}\)
\(\Leftrightarrow A=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{6}\)
\(\Leftrightarrow A=\frac{1}{2-x}\)
Vậy \(A=\frac{1}{2-x}\)
b) \(\left|x\right|=\frac{1}{2}\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\Leftrightarrow x>0\\x=-\frac{1}{2}\Leftrightarrow x< 0\end{matrix}\right.\)
B tự thay vào A rồi tính nhé~
c) A<0
\(\Leftrightarrow\frac{1}{2-x}< 0\Leftrightarrow2-x< 0\Leftrightarrow x>2\)
Vậy với x>2 thì A<0
d) A đạt giá trị nguyên \(\Leftrightarrow\frac{1}{2-x}\in Z\)
Có: \(x\in Z\Rightarrow x-2\in Z\)
\(\Rightarrow x-2\in\text{Ư}\left(1\right)=\left\{\pm1\right\}\)
Bạn tự lập bảng giá trị nhé. lập xong nhớ so sánh ĐKXĐ.