\(9x^2-5x+2\)
\(=\left(3x\right)^2-2.3x.\dfrac{5}{6}+\dfrac{25}{36}+\dfrac{47}{36}\)
\(=\left(3x-\dfrac{5}{6}\right)^2+\dfrac{47}{36}\ge\dfrac{47}{36}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow3x-\dfrac{5}{6}=0\Leftrightarrow3x=\dfrac{5}{6}\Leftrightarrow x=\dfrac{5}{18}\)
Vậy Min của b/t trên là : \(\dfrac{47}{36}\Leftrightarrow x=\dfrac{5}{18}\)
\(2x^2+x+6\)
\(=2\left(x^2+\dfrac{1}{2}x+3\right)\)
\(=2\left(x^2+2x.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{47}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2+\dfrac{47}{16}\right]\)
\(=2\left(x+\dfrac{1}{4}\right)^2+\dfrac{47}{8}\ge\dfrac{47}{8}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x+\dfrac{1}{4}=0\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy Min của b/t trên là : \(\dfrac{47}{8}\Leftrightarrow x=-\dfrac{1}{4}\)
:D