Đáp án:
b. \(m = \dfrac{4}{7}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
y = mx - 2\\
3x + m\left( {mx - 2} \right) = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 2\\
3x + {m^2}x - 2m = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 2\\
\left( {{m^2} + 3} \right)x = 5 + 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 2\\
x = \dfrac{{5 + 2m}}{{{m^2} + 3}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{5 + 2m}}{{{m^2} + 3}}\\
y = m.\dfrac{{5 + 2m}}{{{m^2} + 3}} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{5 + 2m}}{{{m^2} + 3}}\\
y = \dfrac{{5m + 2{m^2}}}{{{m^2} + 3}} - 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{5 + 2m}}{{{m^2} + 3}}\\
y = \dfrac{{5m + 2{m^2} - 2{m^2} - 6}}{{{m^2} + 3}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{5 + 2m}}{{{m^2} + 3}}\\
y = \dfrac{{5m - 6}}{{{m^2} + 3}}
\end{array} \right.\\
Do:{m^2} + 3 \ne 0\forall m \in R
\end{array}\)
⇒ Phương trình có nghiệm với mọi m
\(\begin{array}{l}
b.x + y = 1 - \dfrac{{{m^2}}}{{{m^2} + 3}}\\
\to \dfrac{{5 + 2m}}{{{m^2} + 3}} + \dfrac{{5m - 6}}{{{m^2} + 3}} = \dfrac{{{m^2} + 3 - {m^2}}}{{{m^2} + 3}}\\
\to \dfrac{{7m - 1}}{{{m^2} + 3}} = \dfrac{3}{{{m^2} + 3}}\\
\to 7m - 1 = 3\\
\to m = \dfrac{4}{7}
\end{array}\)
