a)$\begin{array}{l} A = 2{x^2} + {y^2} - 2xy - 2x + 3\\  = {\left( {x - y} \right)^2} + \left( {{x^2} - 2x + 1} \right) + 2\\  = {\left( {x - y} \right)^2} + {\left( {x - 1} \right)^2} + 2 \ge 2\\  \Rightarrow \min A = 2 \Leftrightarrow x = y = 1 \end{array}$

b. Sửa đề $x^2-xy+y^2-3x-3y$

$\begin{array}{l}  \Rightarrow 2B = 2{x^2} - 2xy + 2{y^2} - 6x - 6y\\ 2B = {\left( {x - y} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {x - 3} \right)^2} - 18 \ge  - 18\\  \Rightarrow \min B =  - 9 \Rightarrow x = y = 3\\ c)C = 2{x^2} + 3{y^2} + 4xy - 8x - 2y + 18\\ C = 2{\left( {x + y - 2} \right)^2} + {y^2} + 6y + 9 + 9\\ C = 2{\left( {x + y - 2} \right)^2} + {\left( {y + 3} \right)^2} + 9 \ge 9\\  \Rightarrow \min C = 9 \Rightarrow \left\{ \begin{array}{l} y =  - 4\\ x = 5 \end{array} \right.\\ d)D = 2{x^2} + 3{y^2} + 4{z^2} - 2\left( {x + y + z} \right) + 2\\ D = 2\left( {{x^2} - x + \dfrac{1}{4}} \right) + 3\left( {{y^2} - \dfrac{2}{3}x + \dfrac{1}{9}} \right) + {\left( {2z - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{{12}}\\ D = 2{\left( {x - \dfrac{1}{2}} \right)^2} + 3{\left( {y - \dfrac{1}{3}} \right)^2} + {\left( {2z - \dfrac{1}{2}} \right)^2} + \dfrac{{11}}{2} \ge \dfrac{{11}}{2}\\  \Rightarrow \min D = \dfrac{{11}}{2} \Rightarrow \left\{ \begin{array}{l} x = \dfrac{1}{2}\\ y = \dfrac{1}{3}\\ z = \dfrac{1}{4} \end{array} \right. \end{array}$