Đáp án:
1) \(\left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:m = 2\\
Hpt \to \left\{ \begin{array}{l}
x + y = 2\\
2x + y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x - x + y - y = 3 - 2\\
x + y = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\\
2)\left\{ \begin{array}{l}
\left( {m - 1} \right)x + y = 2\\
mx + y = m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m - 1} \right)x - mx + y - y = 2 - m - 1\\
y = 2 - \left( {m - 1} \right)x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m - 1 - m} \right)x = 1 - m\\
y = 2 - \left( {m - 1} \right)x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m - 1\\
y = 2 - {\left( {m - 1} \right)^2} = 2 - {m^2} + 2m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m - 1\\
y = 1 - {m^2} + 2m
\end{array} \right.\\
Có:2x + y \le 3\\
\to 2m - 2 + 1 - {m^2} + 2m \le 3\\
\to - {m^2} + 4m - 4 \le 0\\
\to - {\left( {m - 2} \right)^2} \le 0\\
\to {\left( {m - 2} \right)^2} \ge 0\left( {ld} \right)\forall m\\
\to dpcm
\end{array}\)
