\(A=x\left(2x-3\right)\\ A=2x^2-3x\\ A=2x^2-3x+\dfrac{9}{8}-\dfrac{9}{8}\\ A=\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{9}{8}\\ A=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\\ A=2\left[x^2-2\cdot x\cdot\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2\right]-\dfrac{9}{8}\\ A=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\\ Do\left(x-\dfrac{3}{4}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{4}\right)^2\ge0\forall x\\ \Rightarrow A=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{3}{4}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{4}=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ \text{ Vậy }A_{\left(Min\right)}=-\dfrac{9}{8}\text{ khi }x=\dfrac{3}{4}\)

\(B=x\left(x-3\right)\\ B=x^2-3x\\ B=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\\ B=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}\\ B=\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{4}\\ B=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\\ Do\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow B=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{ Vậy }B_{\left(Min\right)}=-\dfrac{9}{4}\text{ khi }x=\dfrac{3}{2}\)