Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
9{x^2} + 12x + 4{y^2} + 8y + 8 = 0\\
\Leftrightarrow \left( {9{x^2} + 12x + 4} \right) + \left( {4{y^2} + 8y + 4} \right) = 0\\
\Leftrightarrow \left[ {{{\left( {3x} \right)}^2} + 2.3x.2 + {2^2}} \right] + \left[ {{{\left( {2y} \right)}^2} + 2.2y.2 + {2^2}} \right] = 0\\
\Leftrightarrow {\left( {3x + 2} \right)^2} + {\left( {2y + 2} \right)^2} = 0\\
{\left( {3x + 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {2y + 2} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow {\left( {3x + 2} \right)^2} + {\left( {2y + 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {3x + 2} \right)^2} = 0\\
{\left( {2y + 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x + 2 = 0\\
2y + 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{2}{3}\\
y = - 1
\end{array} \right.\\
c,\\
{x^2} + 4{y^2} + 4x - 4y + 5 = 0\\
\Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left( {4{y^2} - 4y + 1} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 2.x.2 + {2^2}} \right) + \left[ {{{\left( {2y} \right)}^2} - 2.2y.1 + {1^2}} \right] = 0\\
\Leftrightarrow {\left( {x + 2} \right)^2} + {\left( {2y - 1} \right)^2} = 0\\
{\left( {x + 2} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {2y - 1} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow {\left( {x + 2} \right)^2} + {\left( {2y - 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} = 0\\
{\left( {2y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + 2 = 0\\
2y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 2\\
y = \dfrac{1}{2}
\end{array} \right.\\
d,\\
4{x^2} - 12x + {y^2} - 4y + 13 = 0\\
\Leftrightarrow \left( {4{x^2} - 12x + 9} \right) + \left( {{y^2} - 4y + 4} \right) = 0\\
\Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}} \right] + \left[ {{y^2} - 2.y.2 + {2^2}} \right] = 0\\
\Leftrightarrow {\left( {2x - 3} \right)^2} + {\left( {y - 2} \right)^2} = 0\\
{\left( {2x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow {\left( {2x - 3} \right)^2} + {\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {2x - 3} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - 3 = 0\\
y - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = 2
\end{array} \right.
\end{array}\)
Em xem lại đề câu b nhé!
