`a.` Với `x≠-2; +2`
Để `|A|=A` thì `A>0`
`=>` \(\dfrac{x+2}{x-2}>0\)
trường hợp `1:` \(\left\{{}\begin{matrix}x+2>0\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x>2\end{matrix}\right.\Leftrightarrow x>2\)
trường hợp `2:` \(\left\{{}\begin{matrix}x+2< 0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -2\\x< 2\end{matrix}\right.\Leftrightarrow x< -2\)
Vậy \(x>2\) hoặc `x< -2`
`c.` xét phương trình `A=m`
\(\Leftrightarrow\dfrac{x+2}{x-2}=m\\ \Leftrightarrow x+2=m\left(x-2\right)\\ \Leftrightarrow x+2=mx-2m\\ \Leftrightarrow x-mx=-2m-2\\ \Leftrightarrow\left(1-m\right)x=-2m-2\\\)
để phương trình có nghiệm thì `1-m≠0 => m≠1`
b) \(x>2\).
\(\left(x+1\right).A=\left(x+1\right).\dfrac{x+2}{x-2}=\dfrac{x^2+3x+2}{x-2}=\dfrac{x^2-2x+5x-10+12}{x-2}=\dfrac{x\left(x-2\right)+5\left(x-2\right)+12}{x-2}=x+5+\dfrac{12}{x-2}=x-2+\dfrac{12}{x-2}+7\ge2\sqrt{\left(x-2\right).\dfrac{12}{\left(x-2\right)}}+7=2\sqrt{12}+7\)\(\left(x+1\right).A=2\sqrt{12}+7\Leftrightarrow x=2+\sqrt{12}\)
