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