a) \(A=x^2+2y^2=3x-y+6\)
\(A=\left(x^2+3x+\frac{9}{4}\right)+\left(2y^2-y+\frac{1}{8}\right)+\frac{29}{8}\)
\(A=\left(x+\frac{3}{2}\right)^2+\left(\sqrt{2}y-\frac{1}{2\sqrt{2}}\right)^2+\frac{29}{8}\ge\frac{29}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\\sqrt{2}y=\frac{1}{2\sqrt{2}}\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{4}\end{cases}}}\)
Vậy \(Min_A=\frac{29}{8}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{4}\end{cases}}\)
b) \(B=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}\)
Để B min \(\Leftrightarrow\frac{2}{x^2+1}\)max \(\Leftrightarrow x^2+1\)min
Mà \(x^2+1\ge1\)
Dấu " = " xảy ra : \(\Leftrightarrow x=0\)
Vậy \(Min_B=-1\Leftrightarrow x=0\)
