Đáp án:

\(\begin{array}{l}
2,\\
\left\{ \begin{array}{l}
x = \dfrac{5}{2}\\
y = 1
\end{array} \right.\\
4,\\
\left\{ \begin{array}{l}
x =  - \dfrac{1}{2}\\
y = \dfrac{1}{3}
\end{array} \right.\\
6,\\
\left\{ \begin{array}{l}
x = \dfrac{1}{5}\\
y = \dfrac{2}{3}
\end{array} \right.\\
8,\\
\left\{ \begin{array}{l}
x =  - \dfrac{2}{3}\\
y =  - 1
\end{array} \right.\\
10,\\
\left\{ \begin{array}{l}
x =  - 1\\
y = 2
\end{array} \right.\\
12,\\
\left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
y = 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
2,\\
4{x^2} + {y^2} - 20x - 2y + 26 = 0\\
 \Leftrightarrow \left( {4{x^2} - 20x + 25} \right) + \left( {{y^2} - 2y + 1} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} - 2.2x.5 + {5^2}} \right] + \left( {{y^2} - 2.y.1 + {1^2}} \right) = 0\\
 \Leftrightarrow {\left( {2x - 5} \right)^2} + {\left( {y - 1} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
\left. \begin{array}{l}
{\left( {2x - 5} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 1} \right)^2} \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {2x - 5} \right)^2} + {\left( {y - 1} \right)^2} \ge 0,\,\,\,\forall \,x,y\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {2x - 5} \right)^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - 5 = 0\\
y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{5}{2}\\
y = 1
\end{array} \right.\\
4,\\
4{x^2} + 4x - 6y + 9{y^2} + 2 = 0\\
 \Leftrightarrow \left( {4{x^2} + 4x + 1} \right) + \left( {9{y^2} - 6y + 1} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}} \right] + \left[ {{{\left( {3y} \right)}^2} - 2.3y.1 + {1^2}} \right] = 0\\
 \Leftrightarrow {\left( {2x + 1} \right)^2} + {\left( {3y - 1} \right)^2} = 0\,\,\,\,\,\,\,\left( 1 \right)\\
\left. \begin{array}{l}
{\left( {2x + 1} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {3y - 1} \right)^2} \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {2x + 1} \right)^2} + {\left( {3y - 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {2x + 1} \right)^2} = 0\\
{\left( {3y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x + 1 = 0\\
3y - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - \dfrac{1}{2}\\
y = \dfrac{1}{3}
\end{array} \right.\\
6,\\
25{x^2} - 10x + 9{y^2} - 12y + 5 = 0\\
 \Leftrightarrow \left( {25{x^2} - 10x + 1} \right) + \left( {9{y^2} - 12y + 4} \right) = 0\\
 \Leftrightarrow \left[ {{{\left( {5x} \right)}^2} - 2.5x.1 + {1^2}} \right] + \left[ {{{\left( {3y} \right)}^2} - 2.3y.2 + {2^2}} \right] = 0\\
 \Leftrightarrow {\left( {5x - 1} \right)^2} + {\left( {3y - 2} \right)^2} = 0\,\,\,\,\,\,\left( 1 \right)\\
\left. \begin{array}{l}
{\left( {5x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {3y - 2} \right)^2} \ge 0,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {5x - 1} \right)^2} + {\left( {3y - 2} \right)^2} \ge 0,\,\,\forall x,y\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {5x - 1} \right)^2} = 0\\
{\left( {3y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
5x - 1 = 0\\
3y - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{5}\\
y = \dfrac{2}{3}
\end{array} \right.\\
8,\\
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( 1 \right)\\
\left. \begin{array}{l}
{\left( {3x + 2} \right)^2} \ge 0,\,\,\forall x\\
{\left( {2y + 2} \right)^2} \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {3x + 2} \right)^2} + {\left( {2y + 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
\left( 1 \right) \Leftrightarrow \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.\\
10,\\
3{x^2} + 3{y^2} + 6x - 12y + 15 = 0\\
 \Leftrightarrow 3.\left( {{x^2} + {y^2} + 2x - 4y + 5} \right) = 0\\
 \Leftrightarrow {x^2} + {y^2} + 2x - 4y + 5 = 0\\
 \Leftrightarrow \left( {{x^2} + 2x + 1} \right) + \left( {{y^2} - 4y + 4} \right) = 0\\
 \Leftrightarrow \left( {{x^2} + 2.x.1 + {1^2}} \right) + \left( {{y^2} - 2.y.2 + {2^2}} \right) = 0\\
 \Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} = 0\,\,\,\,\,\,\left( 1 \right)\\
\left. \begin{array}{l}
{\left( {x + 1} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {x + 1} \right)^2} + {\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 0\\
y - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
y = 2
\end{array} \right.\\
12,\\
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( 1 \right)\\
\left. \begin{array}{l}
{\left( {2x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\forall y
\end{array} \right\} \Rightarrow {\left( {2x - 3} \right)^2} + {\left( {y - 2} \right)^2} \ge 0,\,\,\,\forall x,y\\
\left( 1 \right) \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}\)