a: Ta có: \(A=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{x+2}\right)\cdot\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)}{x}\)
\(=\dfrac{4x}{x+2}\cdot\dfrac{-1}{x}\)
\(=-\dfrac{4}{x+2}\)
b: Ta có: \(2x^2+x=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=-4:\left(-\dfrac{1}{2}+2\right)=-4:\dfrac{3}{2}=-4\cdot\dfrac{2}{3}=-\dfrac{8}{3}\)
a)\(\left(\dfrac{1}{x-2}-\dfrac{2x}{4-2x}+\dfrac{1}{2+x}\right)\left(\dfrac{2}{x}-1\right)\)
\(=\left[\dfrac{-\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}-\dfrac{2x}{4-2x}+\dfrac{2-x}{\left(2-x\right)\left(2+x\right)}\right].\left(\dfrac{2-x}{x}\right)\)
\(=\left[\dfrac{-\left(x+2\right)-2x+2-x}{4\left(x+2\right)\left(2-x\right)}\right].\dfrac{2-x}{x}\)
\(=\dfrac{-4x}{\left(x+2\right)\left(2-x\right)}.\dfrac{2-x}{x}=-\dfrac{4}{x+2}\)
b)\(2x^2+x=0.x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{1}{2}\Rightarrow A=-\dfrac{8}{3}\end{matrix}\right.\)
c)\(A=\dfrac{1}{2}\Rightarrow-\dfrac{4}{x+2}=\dfrac{1}{2}\Leftrightarrow-8=x+2\)
\(\Leftrightarrow x=-10\)
d)\(A\in Z^+\Leftrightarrow\left\{{}\begin{matrix}x+2\inư\left(-4\right)\\x+2< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+2=-1\\x+2=-4\\x+2=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-6\\x=-4\end{matrix}\right.\)
c: Để \(A=\dfrac{1}{2}\) thì \(\dfrac{-4}{x+2}=\dfrac{1}{2}\)
\(\Leftrightarrow x+2=-8\)
hay x=-10
d: Để A là số nguyên dương thì \(\left\{{}\begin{matrix}x+2\in\left\{1;-1;2;-2;4;-4\right\}\\x>-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{-1;-3;0;-4;2;-6\right\}\\x>-2\end{matrix}\right.\Leftrightarrow x\in\left\{-1;0\right\}\)
