a) Ta có: \(A=2x+3-\sqrt{4x^2-12x+9}\)
\(=2x+3-\sqrt{\left(2x-3\right)^2}\)
\(=2x+3-\left|2x-3\right|\)
\(=\left[{}\begin{matrix}2x+3-2x+3\left(x\ge\frac{3}{2}\right)\\2x+3+2x-3\left(x< \frac{3}{2}\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}6\\4x\end{matrix}\right.\)
b) Vì \(x=\frac{1}{2}< \frac{3}{2}\) nên \(A=4\cdot x=4\cdot\frac{1}{2}=2\)