\(B=4x^2-3x+1\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{1}{4}\right)\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{9}{64}+\dfrac{7}{64}\right)\)
\(=4\left[\left(x-\dfrac{3}{8}\right)^2+\dfrac{7}{64}\right]\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{7}{16}\ge\dfrac{7}{16}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{3}{8}=0\Leftrightarrow x=\dfrac{3}{8}\)
Vậy Min B là : \(\dfrac{7}{16}\Leftrightarrow x=\dfrac{3}{8}\)
B = \(4x^2\) - 3x + 1 = (2x)\(^2\) - 2.2x.\(\dfrac{3}{4}\) + \(\dfrac{9}{16}\) + 1 - \(\dfrac{9}{16}\) = ( 2x - \(\dfrac{3}{4}\) )\(^2\) + \(\dfrac{7}{16}\) \(\ge\) \(\dfrac{7}{16}\)
Vì ( 2x - \(\dfrac{3}{4}\) )\(^2\) \(\ge\) 0 \(\forall\) x
dấu '' = '' xảy ra \(\Leftrightarrow\) :
2x - \(\dfrac{3}{4}\) = 0 \(\Rightarrow\) 2x = \(\dfrac{3}{4}\) \(\Rightarrow\) x = \(\dfrac{3}{8}\)
Vậy GTNN của B = \(\dfrac{7}{16}\) \(\Leftrightarrow\) x = \(\dfrac{3}{8}\)
Ta có:
\(B=4x^2-3x+1\)
B=((2x)2-2.2x.\(\dfrac{3}{4}\)+\(\dfrac{9}{16}\))+\(\dfrac{7}{16}\)
B=\(\left(2x-\dfrac{3}{4}\right)^2+\dfrac{7}{16}\)
Xét \(\left(2x-\dfrac{3}{4}\right)^2\ge0\)
=> \(\left(2x-\dfrac{3}{4}\right)^2+\dfrac{7}{16}\ge\dfrac{7}{16}\)
Vậy min B là \(\dfrac{7}{16}\) khi \(2x-\dfrac{3}{4}=0=>x=\dfrac{3}{8}\)
