Đáp án:
h. \(\left\{ \begin{array}{l}
y = 5\\
x = 6
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
g.\left\{ \begin{array}{l}
x = 5 - 2y\\
{\left( {5 - 2y + 3} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {5 - 2y} \right)^2} + {y^2}\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to {\left( {8 - 2y} \right)^2} + {y^2} - 2y + 1 = 25 - 20y + 4{y^2} + {y^2}\\
\to 64 - 32y + 4{y^2} + {y^2} - 2y + 1 = 25 - 20y + 4{y^2} + {y^2}\\
\to 14y = 40\\
\to y = \dfrac{{20}}{7}\\
\to x = - \dfrac{5}{7}\\
i.\left\{ \begin{array}{l}
y = 5 - 3x\\
{\left( {x - 1} \right)^2} - 5 + 3x + y = {x^2} + 10\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) \to {x^2} - 2x + 1 - 5 + 3x + y = {x^2} + 10\\
\to x = 14\\
\to y = - 37\\
h.\left\{ \begin{array}{l}
xy - y + x - 1 = xy\\
2x - 3y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - y = 1\\
2x - 3y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 + y\\
2\left( {1 + y} \right) - 3y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 + y\\
2 + 2y - 3y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 5\\
x = 6
\end{array} \right.
\end{array}\)
