\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
a) ĐKXD: \(x+2\ne0\)và \(x^2+4x+4\ne0\)và \(x^2-4\ne0\)và \(2-x\ne0\)
\(\Leftrightarrow x\ne-2\)và \(\left(x+2\right)^2\ne0\)và \(\left(x-2\right)\left(x+2\right)\ne0\)và \(x\ne2\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
+) \(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=\frac{-2x+4}{x+2}\)
b) Ta có: x-1=3 <=> x=4 Thay vào A ta được:
\(\frac{-2.4-4}{4+2}=-2\)
c)
Để \(A\in Z\Leftrightarrow8⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(8\right)=\left\{\pm1;\pm4;\pm8\right\}\)
Bạn làm nốt nha
