Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
D = 2{x^2} + 9{y^2} - 6xy - 6x + 12y + 2012\\
= \left( {{x^2} - 6xy + 9{y^2}} \right) - 4.\left( {x - 3y} \right) + \left( {{x^2} - 2x + 1} \right) + 2011\\
= {\left( {x - 3y} \right)^2} - 4.\left( {x - 3y} \right) + {\left( {x - 1} \right)^2} + 2011\\
= \left[ {{{\left( {x - 3y} \right)}^2} - 4.\left( {x - 3y} \right) + 4} \right] + {\left( {x - 1} \right)^2} + 2007\\
= {\left( {x - 3y - 2} \right)^2} + {\left( {x - 1} \right)^2} + 2007 \ge 2007,\,\,\forall x,y\\
\Rightarrow {D_{\min }} = 2007 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 3y - 2} \right)^2} = 0\\
{\left( {x - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = - \frac{1}{3}
\end{array} \right.\\
*)\\
E = {x^2} - 2xy + 4{y^2} - 2x + 10y + 29\\
= \left( {{x^2} - 2xy + {y^2}} \right) - 2.\left( {x - y} \right) + \left( {3{y^2} + 8y + \frac{{16}}{3}} \right) + \frac{{71}}{3}\\
= {\left( {x - y} \right)^2} - 2\left( {x - y} \right) + 1 + 3.\left( {{y^2} + \frac{8}{3}y + \frac{{16}}{9}} \right) + \frac{{68}}{3}\\
= {\left( {x - y} \right)^2} - 2.\left( {x - y} \right) + 1 + 3.{\left( {y + \frac{4}{3}} \right)^2} + \frac{{68}}{3}\\
= {\left( {x - y - 1} \right)^2} + 3.{\left( {y + \frac{4}{3}} \right)^2} + \frac{{68}}{3} \ge \frac{{68}}{3},\,\,\,\forall x,y\\
\Rightarrow {E_{\min }} = \frac{{68}}{3} \Leftrightarrow \left\{ \begin{array}{l}
x - y - 1 = 0\\
y + \frac{4}{3} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \frac{1}{3}\\
y = - \frac{4}{3}
\end{array} \right.\\
*)\\
F = \frac{3}{{2x - {x^2} - 4}}\\
2x - {x^2} - 4 = - \left( {{x^2} - 2x + 1} \right) - 3 = - {\left( {x - 1} \right)^2} - 3 \le - 3,\,\,\,\forall x\\
\Rightarrow F = \frac{3}{{ - {{\left( {x - 1} \right)}^2} - 3}} \ge \frac{3}{{ - 3}} = - 1\\
\Rightarrow {F_{\min }} = - 1 \Leftrightarrow x = 1\\
*)\\
G = \frac{2}{{6x - 5 - 9{x^2}}}\\
6x - 5 - 9{x^2} = - \left( {9{x^2} - 6x + 1} \right) - 4 = - 4 - {\left( {3x - 1} \right)^2} \le - 4\\
\Rightarrow G \ge \frac{2}{{ - 4}} = - \frac{1}{2}\\
\Rightarrow {G_{\min }} = - \frac{1}{2} \Leftrightarrow x = \frac{1}{3}
\end{array}\)
