\(Q=x^2+2y^2+4x+6y+1\)
\(Q=\left(x^2+4x+4\right)+2\left(y^2+3y+\frac{9}{4}\right)-\frac{15}{2}\)
\(Q=\left(x+2\right)^2+2\left(y+\frac{3}{2}\right)^2-\frac{15}{2}\ge-\frac{15}{2}\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x=-2\\y=-\frac{3}{2}\end{cases}}\)