Bài 1:
\(A=3x^2+2x-3=3(x^2+\frac{2}{3}x+\frac{1}{3^2})-\frac{10}{3}\)
\(=3(x+\frac{1}{3})^2-\frac{10}{3}\geq 3.0-\frac{10}{3}=-\frac{10}{3}\)
Vậy GTNN của $A$ là \(\frac{-10}{3}\).
Dấu "=" xảy ra khi \((x+\frac{1}{3})^2=0\Leftrightarrow x=-\frac{1}{3}\)
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\(B=3x^2-6xy+5y^2-y+3x+2016\)
\(=3(x^2-2xy+y^2)+2y^2-y+3x+2016\)
\(=3(x-y)^2+3(x-y)+2y^2+2y+2016\)
\(=3(x-y)^2+3(x-y)+\frac{3}{4}+2(y^2+y+\frac{1}{4})+\frac{8059}{4}\)
\(=3[(x-y)^2+(x-y)+\frac{1}{4}]+2(y+\frac{1}{2})^2+\frac{8059}{4}\)
\(=3(x-y+\frac{1}{2})^2+2(y+\frac{1}{2})^2+\frac{8059}{4}\)
\(\geq 3.0+2.0+\frac{8059}{4}=\frac{8059}{4}\)
Vậy GTNN của $B$ là \(\frac{8059}{4}\).
Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-y+\frac{1}{2}=0\\ y+\frac{1}{2}=0\end{matrix}\right.\Leftrightarrow x=-1; y=-\frac{1}{2}\)