Đáp án:
a) \(\left\{ \begin{array}{l}
x = \dfrac{{11}}{6}\\
y = \dfrac{5}{6}
\end{array} \right.\)
b) \(m \in \emptyset \)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 5\\
Hpt \to \left\{ \begin{array}{l}
x + 5y = 6\\
5x + y = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + 5y = 6\\
- 25x - 5y = - 50
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 24x = - 44\\
x + 5y = 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{11}}{6}\\
y = \dfrac{5}{6}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
x + my = m + 1\\
mx + y = 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
mx + {m^2}y = {m^2} + m\\
mx + y = 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} - 1} \right)y = {m^2} - m\\
x = m + 1 - my
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m\left( {m - 1} \right)}}{{\left( {m - 1} \right)\left( {m + 1} \right)}}\\
x = m + 1 - my
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{m}{{m + 1}}\\
x = m + 1 - m.\dfrac{m}{{m + 1}} = \dfrac{{{m^2} + 2m + 1 - {m^2}}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{m}{{m + 1}}\\
x = \dfrac{{2m + 1}}{{m + 1}}
\end{array} \right.\\
DK:m \ne \pm 1\\
x + 5y = 4\\
\to \dfrac{{2m + 1}}{{m + 1}} + 5.\dfrac{m}{{m + 1}} = 4\\
\to \dfrac{{2m + 1 + 5m - 4m - 4}}{{m + 1}} = 0\\
\to 3m - 3 = 0\\
\to m = 1\left( {KTM} \right)\\
\to m \in \emptyset
\end{array}\)
