\(A=3x^2-2x-1\)
\(A=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)
\(A=3\left(x^2-2.\frac{1}{3}.x+\frac{1}{9}-\frac{4}{9}\right)\)
\(A=3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\ge-\frac{4}{3}\)với \(\forall x\)
Dấu "=" xảy ra <=> x=1/3
Vậy GTNN của A là -4/3 <=> x=1/3
\(A=3x^2-2x-1\)
\(A=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)
\(A=3\left(x^2-\frac{2}{3}x+\frac{1}{9}-\frac{4}{9}\right)\)
\(A=3\left(x-\frac{1}{3}\right)^2-\frac{12}{9}\ge-\frac{12}{9}\)
Dấu '' = '' xảy ra
\(\Leftrightarrow3\left(x-\frac{1}{3}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{3}=0\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy GTNN của A \(\Leftrightarrow x=\frac{1}{3}\)
