\(A=x\left(2x-3\right)\\ A=2x^2-3x\\ A=2x^2-3x+\dfrac{9}{8}-\dfrac{9}{8}\\ A=\left(2x^2-3x+\dfrac{9}{8}\right)-\dfrac{9}{8}\\ A=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\\ A=2\left[x^2-2\cdot x\cdot\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2\right]-\dfrac{9}{8}\\ A=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\\ Do\left(x-\dfrac{3}{4}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{4}\right)^2\ge0\forall x\\ \Rightarrow A=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{3}{4}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{4}=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ \text{ Vậy }A_{\left(Min\right)}=-\dfrac{9}{8}\text{ khi }x=\dfrac{3}{4}\)
\(B=x\left(x-3\right)\\ B=x^2-3x\\ B=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\\ B=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}\\ B=\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{9}{4}\\ B=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\\ Do\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow B=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{ Vậy }B_{\left(Min\right)}=-\dfrac{9}{4}\text{ khi }x=\dfrac{3}{2}\)
a, A=x( 2x-3)
=2x2 -3x
= 2(x2 - \(\dfrac{3}{2}\)x)
= 2(x2-2.\(\dfrac{3}{4}\)x +\(\dfrac{9}{16}+\dfrac{7}{16}\) )
= 2(x-\(\dfrac{3}{4}\))2 +\(\dfrac{7}{16}\)
mà: 2(x-\(\dfrac{3}{4}\))2 \(\ge0\forall x\in R\)
=> 2(x-\(\dfrac{3}{4}\))2 +\(\dfrac{7}{16}\) \(\ge\dfrac{7}{16}\forall x\in R\)
=> Min A= 7/16
