Đáp án:
a. \(\left\{ \begin{array}{l}
x = 1\\
y = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left\{ \begin{array}{l}
3x + y = 2\\
2x - 3y = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
9x + 3y = 6\\
2x - 3y = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
11x = 11\\
2x - 3y = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = - 1
\end{array} \right.\\
b.\left\{ \begin{array}{l}
{x^2} + 2y + 1 = 0\\
{y^2} + 2x + 1 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x^2} - {y^2} + 2\left( {x - y} \right) = 0\\
{x^2} + 2y + 1 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x - y} \right)\left( {x + y} \right) + 2\left( {x - y} \right) = 0\\
{x^2} + 2y + 1 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {x - y} \right)\left( {x + y + 2} \right) = 0\\
{x^2} + 2y + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = y\\
{x^2} + 2y + 1 = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - x - 2\\
{x^2} + 2y + 1 = 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = y\\
{x^2} + 2x + 1 = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - 2x - 4 + 1 = 0\\
y = - x - 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = y\\
{\left( {x + 1} \right)^2} = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
y = - x - 2\\
\left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = y = - 1\\
x = 3 \to y = - 5
\end{array} \right.
\end{array}\)
