Question

tìm GTLN hoặc GTNN của đa thức sau

B=\(4x^2-3x+1\)

giúp mình với mình đang cần gấp

Answer 1

\(B=4x^2-3x+1\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{1}{4}\right)\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{9}{64}+\dfrac{7}{64}\right)\)

\(=4\left[\left(x-\dfrac{3}{8}\right)^2+\dfrac{7}{64}\right]\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{7}{16}\ge\dfrac{7}{16}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{3}{8}=0\Leftrightarrow x=\dfrac{3}{8}\)

Vậy Min B là : \(\dfrac{7}{16}\Leftrightarrow x=\dfrac{3}{8}\)

Answer 2

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 = \(\dfrac{3}{8}\)

Answer 3

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 \(2x-\dfrac{3}{4}=0=>x=\dfrac{3}{8}\)