Đáp án:
a) m=-3
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
\left\{ \begin{array}{l}
x = my + 3\\
m\left( {my + 3} \right) - 4y = m + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = my + 3\\
{m^2}y + 3m - 4y = m + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = my + 3\\
\left( {{m^2} - 4} \right)y = - 2m + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = my + 3\\
y = \dfrac{{ - 2\left( {m - 2} \right)}}{{\left( {m - 2} \right)\left( {m + 2} \right)}} = - \dfrac{2}{{m + 2}}
\end{array} \right.\\
\to x = m.\left( { - \dfrac{2}{{m + 2}}} \right) + 3\\
= \dfrac{{ - 2m + 3m + 6}}{{m + 2}} = \dfrac{{m + 6}}{{m + 2}}\\
DK:m \ne \pm 2\\
Do:x + y = \dfrac{3}{m}\left( {m \ne 0} \right)\\
\to \dfrac{{m + 6}}{{m + 2}} - \dfrac{2}{{m + 2}} = \dfrac{3}{m}\\
\to \dfrac{{m + 4}}{{m + 2}} = \dfrac{3}{m}\\
\to {m^2} + 4m = 3m + 6\\
\to {m^2} + m - 6 = 0\\
\to \left( {m + 3} \right)\left( {m - 2} \right) = 0\\
\to \left[ \begin{array}{l}
m = - 3\left( {TM} \right)\\
m = 2\left( l \right)
\end{array} \right.\\
b)Do:x > 0;y < 0\\
\to \left\{ \begin{array}{l}
\dfrac{{m + 6}}{{m + 2}} > 0\\
- \dfrac{2}{{m + 2}} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m + 2 > 0\\
m + 6 > 0
\end{array} \right.\\
\to m > - 2;m \ne 2
\end{array}\)
