a) Ta có: \(A=\left(\frac{3}{2x+4}+\frac{x}{2-x}+\frac{2x^2+3}{x^2-4}\right):\frac{2x-1}{4x-8}\)
\(=\left(\frac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\frac{2x\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{4\left(x-2\right)}{2x-1}\)
\(=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}\)
\(=\frac{2x^2-x}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}\)
\(=\frac{x\left(2x-1\right)\cdot4\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)\cdot\left(2x-1\right)}\)
\(=\frac{2x}{x+2}\)
b)
ĐKXĐ: \(x\notin\left\{2;-2;\frac{1}{2}\right\}\)
Để A<2 thì A-2<0
hay \(\frac{2x}{x+2}-2< 0\)
\(\Leftrightarrow\frac{2x}{x+2}-\frac{2\left(x+2\right)}{x+2}< 0\)
\(\Leftrightarrow\frac{2x-2x-4}{x+2}< 0\)
\(\Leftrightarrow\frac{-4}{x+2}< 0\)
\(\Leftrightarrow-4;x+2\) khác dấu
mà -4<0
nên x+2>0
hay x>-2
mà \(x\notin\left\{2;-2;\frac{1}{2}\right\}\)
nên \(\left\{{}\begin{matrix}x>-2\\x\notin\left\{\frac{1}{2};2\right\}\end{matrix}\right.\)
Vậy: Để A<2 thì \(\left\{{}\begin{matrix}x>-2\\x\notin\left\{\frac{1}{2};2\right\}\end{matrix}\right.\)
c) Ta có: |x-1|=3
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Thay x=4 vào biểu thức \(A=\frac{2x}{x+2}\), ta được:
\(\frac{2\cdot4}{4+2}=\frac{8}{6}=\frac{4}{3}\)
Vậy: \(\frac{4}{3}\) là giá trị của biểu thức \(A=\frac{2x}{x+2}\) tại x=4
d) Để |A|=1 thì
\(\left[{}\begin{matrix}A=1\\A=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2x}{x+2}=1\\\frac{2x}{x+2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=x+2\\2x=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=2\\2x+x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\3x=-2\end{matrix}\right.\Leftrightarrow x=\frac{-2}{3}\)
Vậy: Để |A|=1 thì \(x=\frac{-2}{3}\)