\(A=3x^2-12x+10\\ A=3x^2-12x+12-2\\ A=\left(3x^2-12x+12\right)-2\\ A=3\left(x^2-4x+4\right)-2\\ A=3\left(x^2-2\cdot x\cdot2+2^2\right)-2\\ A=3\left(x-2\right)^2-2\\ Do\left(x-2\right)^2\ge0\forall x\\ \Rightarrow3\left(x-2\right)^2\ge0\forall x\\ \Rightarrow A=3\left(x-2\right)^2-2\ge-2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{ Vậy }A_{\left(Min\right)}=-2\text{ khi }x=2\)
A=3x2 - 12x + 10
A= (3x2- 2.3x.2+22)-22+10
A= (3x-2)2+6 \(\ge\) +6
Vậy min A = 6 . Dấu = xảy ra khi 3x -2 = 0
3x= 2
x= \(\dfrac{2}{3}\)
A = \(3x^2-12x+10\)
\(=\left(3x^2-12x+12\right)-12+10\)
\(=\left(\sqrt{3}x\right)^2+2.\sqrt{3}x.2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+10\)
\(=\left(\sqrt{3}x+2\sqrt{3}\right)^2-2\)
Ta có :
\(\left(\sqrt{3}x+2\sqrt{3}\right)^2\ge0\forall x\)
\(\Rightarrow\) \(\left(\sqrt{3}x+2\sqrt{3}\right)^2-2\ge0\forall x\)
\(\Rightarrow A\ge0\)
Dấu = xảy ra khi
\(\sqrt{3}x+2\sqrt{3}=0\)
\(\Rightarrow\sqrt{3}x=-2\sqrt{3}\)
\(\Rightarrow x=-2\)
Vậy \(Min_A=0\Leftrightarrow x=-2\)
