Đáp án:
a) \(\left[ \begin{array}{l}
m = 0\\
m = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
y = mx - 2\\
x + m\left( {mx - 2} \right) = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + {m^2}x - 2m = 1\\
y = mx - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {1 + {m^2}} \right)x = 2m + 1\\
y = mx - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2m + 1}}{{1 + {m^2}}}\\
y = m.\dfrac{{2m + 1}}{{1 + {m^2}}} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2m + 1}}{{1 + {m^2}}}\\
y = \dfrac{{2{m^2} + m - 2{m^2} - 2}}{{1 + {m^2}}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2m + 1}}{{1 + {m^2}}}\\
y = \dfrac{{m - 2}}{{1 + {m^2}}}
\end{array} \right.\\
a)x + y = - 1\\
\to \dfrac{{2m + 1}}{{1 + {m^2}}} + \dfrac{{m - 2}}{{1 + {m^2}}} = - 1\\
\to 3m - 1 = - {m^2} - 1\\
\to {m^2} + 3m = 0\\
\to m\left( {m + 3} \right) = 0\\
\to \left[ \begin{array}{l}
m = 0\\
m = - 3
\end{array} \right.\\
\end{array}\)
\(\begin{array}{l}
b)\left\{ \begin{array}{l}
m = \dfrac{{y + 2}}{x}\\
x + \dfrac{{y + 2}}{x}.y = 1\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to \dfrac{{{x^2} + {y^2} + 3y - x}}{x} = 0
\end{array}\)
\(\left( {DK:x \ne 0} \right)\)
