\(a,A=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\left(\dfrac{2x-1}{4x-8}\right)\)
\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{-2x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right):\dfrac{2x-1}{4\left(x-2\right)}\)
\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}.\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x}{x+2}.\dfrac{2}{2x-1}\)
\(=\dfrac{2x}{x+2}\)
\(b,\) Để A < 2
\(\Leftrightarrow\dfrac{2x}{x+2}< 2\)
\(\Leftrightarrow\dfrac{2x}{x+2}-\dfrac{2\left(x+2\right)}{x+2}< 0\)
\(\Leftrightarrow\dfrac{-4}{x+2}< 0\)
\(\Leftrightarrow x+2>0\)
\(\Leftrightarrow x>-2\)
Vậy...............................
c,ĐKXĐ của A : x ≠ -2
\(\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(t/m\right)\\x=-2\left(L\right)\end{matrix}\right.\)
Thay x = 4 vào bt A : \(\dfrac{2.4}{4+2}=\dfrac{8}{6}=\dfrac{4}{3}\)
Vậy..................