\(A=9x^2+2x-1\)
vì \(9x^2\)luôn \(\ge0\)\(\forall x\in Q\)
\(\Rightarrow9x^2+2x-1\)\(\ge2x-1\)\(\forall x\in Q\)
dấu ''='' xảy ra <=> \(9x^2=0\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
vậy gtnn của bt a là 2x-1 tại x=0
\(A=9x^2+2x-1\)
\(=\left(3x\right)^2+2.3x.\frac{1}{3}+\frac{1}{9}-\frac{10}{9}\)
\(=\left(3x+\frac{1}{3}\right)^2-\frac{10}{9}\ge-\frac{10}{9}\)
Vậy \(A_{min}=-\frac{10}{9}\Leftrightarrow3x+\frac{1}{3}=0\Leftrightarrow x=\frac{-1}{9}\)
