Đáp án:
\[x = 1;\,\,y = - 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
10{x^2} + 10{y^2} + 16xy + 4y - 4x + 4 = 0\\
\Leftrightarrow 2.\left( {5{x^2} + 5{y^2} + 8xy + 2y - 2x + 2} \right) = 0\\
\Leftrightarrow 5{x^2} + 5{y^2} + 8xy + 2y - 2x + 2 = 0\\
\Leftrightarrow \left( {4{x^2} + 8xy + 4{y^2}} \right) + \left( {{x^2} - 2x + 1} \right) + \left( {{y^2} + 2y + 1} \right) = 0\\
\Leftrightarrow \left[ {{{\left( {2x} \right)}^2} + 2.2x.2y + {{\left( {2y} \right)}^2}} \right] + \left( {{x^2} - 2.x.1 + {1^2}} \right) + \left( {{y^2} + 2.y.1 + {1^2}} \right) = 0\\
\Leftrightarrow {\left( {2x + 2y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 0\,\,\,\,\,\left( 1 \right)\\
{\left( {2x + 2y} \right)^2} \ge 0,\,\,\,\forall \,x,y\\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall \,x\\
{\left( {y + 1} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow {\left( {2x + 2y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} \ge 0,\,\,\,\forall \,x,y\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {2x + 2y} \right)^2} = 0\\
{\left( {x - 1} \right)^2} = 0\\
{\left( {y + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x + 2y = 0\\
x - 1 = 0\\
y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = - 1
\end{array} \right.\\
Vậy\,\,x = 1;\,\,y = - 1
\end{array}\)
