a, đk x khác 1; -2 ; 2
\(B=\left(\dfrac{2x\left(x+2\right)-x^2-5x}{x^2-4}\right):\left(\dfrac{3x+3}{x+2}-2\right)\)
\(=\left(\dfrac{x^2-x}{x^2-4}\right):\left(\dfrac{3x+3-2x-4}{x+2}\right)=\dfrac{x\left(x-1\right)\left(x+2\right)}{\left(x^2-4\right)\left(x-1\right)}=\dfrac{x}{x-2}\)
b, \(\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-1\end{matrix}\right.\)
Thay vào ta được -1/(-1-2) = -1/-3 = 1/3
c, Ta có \(\dfrac{x}{x-2}-1< 0\Leftrightarrow\dfrac{x-x+2}{x-2}< 0\Leftrightarrow\left\{{}\begin{matrix}x-2\ne0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x< 2\end{matrix}\right.\)
Kết hợp đk vậy x < 2 ; x khác 1 ; -2
