Bài 2:
a: Ta có: \(A=3x^2+x+2\)
\(=3\left(x^2+\dfrac{1}{3}x+\dfrac{2}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)
\(=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{6}\)
