b) Ta có: A=x(2x-3)
\(=2x^2-3x\)
\(=2\left(x^2-\dfrac{3}{2}x\right)\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}\right)\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{4}\)