Đáp án:
\(\left[ \begin{array}{l}
m = \dfrac{{1 + \sqrt {17} }}{4}\\
m = \dfrac{{1 - \sqrt {17} }}{4}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
6b.\left\{ \begin{array}{l}
y = mx - 1\\
2x + {m^2}x - m = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 1\\
\left( {{m^2} + 2} \right)x = m + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = mx - 1\\
x = \dfrac{{m + 4}}{{{m^2} + 2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 4}}{{{m^2} + 2}}\\
y = \dfrac{{{m^2} + 4 - {m^2} - 2}}{{{m^2} + 2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m + 4}}{{{m^2} + 2}}\\
y = \dfrac{2}{{{m^2} + 2}}
\end{array} \right.\\
Có:x + y = 2\\
\to \dfrac{{m + 4}}{{{m^2} + 2}} + \dfrac{2}{{{m^2} + 2}} = 2\\
\to m + 6 = 2{m^2} + 4\\
\to 2{m^2} - m - 2 = 0\\
Δ= 1 - 4.2.\left( { - 2} \right) = 17\\
\to \left[ \begin{array}{l}
m = \dfrac{{1 + \sqrt {17} }}{4}\\
m = \dfrac{{1 - \sqrt {17} }}{4}
\end{array} \right.
\end{array}\)
