\(A=4x^2-8x+2017\)
\(=\left(4x^2-8x+4\right)+2013\)
\(=4\left(x^2-2x+1\right)+2013\)
\(=4\left(x-1\right)^2+2013\)
Ta có :
\(4\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow4\left(x-1\right)^2+2013\ge2013\forall x\)
Dấu = xảy ra \(\Leftrightarrow4\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy \(Min_A=2013\Leftrightarrow x=1\)
A = 4x2 - 8x + 2017
A=4x2-8x+4+2013
A=(4x2-8x+4)+2013
A=4(x2-2x+1)+2013
A=4(x-1)2+2013
Do (x-1)2 ≥ 0 ∀ x
=>4(x-1)2≥0
=>4(x-1)2+2013≥2013
=>A≥2013
=>MinA=2013 khi
x-1=0
=>x=1
vậy MinA =2013 khi và chỉ khi x=1
A= 4x2 - 8x + 2017
A= 4x2 - 8x + 4 + 2013
A= (4x2 - 8x + 4) + 3013
A= (2x - 2)2 + 2013
Do (2x - 2)2 \(\ge\) 0 \(\forall\) x
=> (2x - 2)2 + 2013 \(\ge\) 2013 > 0 \(\forall\) x
Dấu " = " xảy ra khi :
(2x - 2)2 = 0
=> 2x - 2 = 0
=> 2x = 2
=> x = 1
Vậy \(\min\limits_{ }\) A = 2013 khi x = 1
