$A=x^2+3xy+6x+5y^2+7y-2$

$=\left[x^2+2x\left(3+\dfrac{3}{2}y\right)+\left(3+\dfrac{3}{2}y\right)^2\right]+5y^2+7y-2-\left(3+\dfrac{3}{2}y\right)^2$$=\left(x+3+\dfrac{3}{2}y\right)^2+5y^2+7y-2-9-9y-\dfrac{9}{4}y^2$$=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}y^2-2y-11$

$=\left(x+3+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\left(y^2-\dfrac{8}{11}y+\dfrac{16}{121}\right)-\dfrac{125}{11}$$=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}\left(x-\dfrac{4}{11}\right)^2-\dfrac{125}{11}\ge\dfrac{-125}{11}$Vậy $Min_A=\dfrac{-125}{11}$ khi $\left[{}\begin{matrix}x+3+\dfrac{3}{2}y=0\\x-\dfrac{4}{11}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{74}{33}\\x=\dfrac{4}{11}\end{matrix}\right.$

Biết số nhọ nhưng vẫn làm tiếp:)