Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.\left\{ \begin{array}{l}
2x + y = 4\\
x - 2y = 2
\end{array} \right. \to \left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.\\
b.\left\{ \begin{array}{l}
2x + y = 5m - 1\\
x - 2y = 2\\
{x^2} - 2{y^2} = 1
\end{array} \right. \to \left\{ \begin{array}{l}
2x + y = 5m\\
x = 2 + 2y\\
4 + 8y + 4{y^2} - 2{y^2} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x + y = 5m\\
x = 2 + 2y\\
\left[ \begin{array}{l}
y = \frac{{ - 4 + \sqrt {10} }}{2}\\
y = \frac{{ - 4 - \sqrt {10} }}{2}
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y = \frac{{ - 4 + \sqrt {10} }}{2}\\
x = - 2 + \sqrt {10} \\
m = \frac{{ - 2 + \sqrt {10} }}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
y = \frac{{ - 4 - \sqrt {10} }}{2}\\
x = - 2 + \sqrt {10} \\
m = \frac{{ - 2 - \sqrt {10} }}{2}
\end{array} \right.
\end{array} \right.
\end{array}\)
