Ta có: \(A=9x^2+\frac{6}{5}x+9\Leftrightarrow A=3x.3x+\frac{3}{5}x+\frac{3}{5}x+\frac{9}{225}+\frac{2016}{225}\)
\(\Leftrightarrow A=3x.3x+3x.\frac{3}{15}+\frac{3}{15}.3x+\frac{3}{15}.\frac{3}{15}+\frac{2016}{225}\)
\(\Leftrightarrow A=3x\left(3x+\frac{3}{15}\right)+\frac{3}{15}\left(3x+\frac{3}{15}\right)+\frac{2016}{225}=\left(3x+\frac{3}{15}\right)\left(3x+\frac{3}{15}\right)+\frac{2016}{225}=\left(3x+\frac{3}{15}\right)^2+\frac{2016}{225}\)
Do \(\left(3x+\frac{3}{15}\right)^2\ge0\Rightarrow\left(3x+\frac{3}{15}\right)^2+\frac{2016}{225}\ge\frac{2016}{225}\Leftrightarrow A\ge\frac{2016}{225}\)
Dấu "=" xảy ra khi: \(\left(3x+\frac{3}{15}\right)^2=0\Leftrightarrow3x+\frac{3}{15}=0\Leftrightarrow3x=-\frac{3}{15}\Leftrightarrow x=-\frac{1}{15}\)
Vậy GTNN của biểu thức \(A\)là \(\frac{2016}{225}\)tại \(x=-\frac{1}{15}.\)
\(Â=9\left(x^2+\frac{2}{15}x\right)+9=9\left(x^2+2xxx\frac{1}{15}+\frac{1}{15^2}\right)+9-9x\frac{1}{15^2}\\ =9\left(x+\frac{1}{15}\right)^2+\frac{224}{25}\)
A >= 224/25
Dấu bằng xảy ra khi và chỉ khi x = -1/5
