Đáp án:
a) \(\left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
C9:\\
a)Thay:m = 2\\
Hpt \to \left\{ \begin{array}{l}
x + y = 2\\
2x + y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 2 - x\\
2x + 2 - x = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\\
b)\left\{ \begin{array}{l}
\left( {m - 1} \right)x + y = 2\\
mx + y = m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m - 1 - m} \right)x = 2 - m - 1\\
mx + y = m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{1 - m}}{{ - 1}} = m - 1\\
y = m + 1 - m\left( {m - 1} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m - 1\\
y = m + 1 - {m^2} + m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = m - 1\\
y = - {m^2} + 2m + 1
\end{array} \right.\\
Có:2x + y = 2\left( {m - 1} \right) - {m^2} + 2m + 1\\
= 2m - 2 - {m^2} + 2m + 1\\
= - {m^2} + 4m - 1\\
= - \left( {{m^2} - 4m + 1} \right)\\
= - \left( {{m^2} - 4m + 4 - 3} \right)\\
= - {\left( {m - 2} \right)^2} + 3\\
Do:{\left( {m - 2} \right)^2} \ge 0\forall x\\
\to - {\left( {m - 2} \right)^2} \le 0\\
\to - {\left( {m - 2} \right)^2} + 3 \le 3
\end{array}\)
⇒ Điều phải chứng minh
