\(4x^2-4x+1-\left(2x-1\right)b+\dfrac{b^2}{4}\)
\(=\left(2x-1\right)^2-2.\left(2x-1\right).\left(\dfrac{1}{2}b\right)+\left(\dfrac{b}{2}\right)^2\)
\(=\left(2x-1-\dfrac{1}{2}b\right)^2\)
Ta có: \(\left(4x^2-4x+1\right)-\left(2x-1\right)b+\dfrac{b^2}{4}\)
\(=\left(2x-1\right)^2-2\cdot\left(2x-1\right)\cdot\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2\)
\(=\left(2x-\dfrac{1}{2}b-1\right)^2\)