\(9x^2+9x+2\\=(3x)^2+2.3x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}\\=\left(3x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=0\)
Vì \(\left(3x+\dfrac{3}{2}\right)^2\ge 0→\left(3x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}≥-\dfrac{1}{4}\)
\(→\min =-\dfrac{1}{4}\)
\(→\) Dấu "=" xảy ra khi \(3x+\dfrac{3}{2}=0\)
\(↔3x=-\dfrac{3}{2}\\↔x=-\dfrac{1}{2}\)
Vậy \(\min =-\dfrac{1}{4}\) khi \(x=-\dfrac{1}{2}\)
