A = 16x2 - 8x + 5
A = [ ( 4x )2 - 2 . 4x . 1 + 1 ] + 4
A = ( 4x - 1 )2 + 4
Vì ( 4x - 1 )2 \(\ge\) 0 \(\forall\)x
=> ( 4x - 1 )2 + 4 \(\ge\)4 \(\forall\)x
=> A \(\ge\)4 \(\forall\)x
=> A = 4 <=> ( 4x - 1 )2 = 0
<=> 4x - 1 = 0
<=> 4x = 1
<=> x = \(\frac{1}{4}\)
Vậy A min = 4 <=> x = \(\frac{1}{4}\)