Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
3x + my = m\\
\left( {m - 1} \right)x + 2y = m - 1\\
x + {y^2} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - {y^2}\\
3 - 3{y^2} + my = m\\
m - 1 - m{y^2} + {y^2} + 2y = m - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - {y^2}\\
3{y^2} - my + m - 3 = 0\\
\left( {1 - m} \right){y^2} + 2y = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - {y^2}\\
\left[ \begin{array}{l}
y = 0\\
y = \frac{2}{{m - 1}}
\end{array} \right. \to \left[ \begin{array}{l}
3.0 - m.0 + m - 3 = 0\\
\frac{{12}}{{{{\left( {m - 1} \right)}^2}}} - \frac{{2m}}{{m - 1}} + m - 3 = 0
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1 - {y^2}\\
\left[ \begin{array}{l}
y = 0 \to m = 3 \to x = 1\\
y = \frac{2}{{m - 1}} \to 12 - 2m\left( {m - 1} \right) + \left( {m - 3} \right)\left( {{m^2} - 2m + 1} \right) = 0\left( * \right)
\end{array} \right.
\end{array} \right.\\
\left( * \right) \to 12 - 2{m^2} + 2m + {m^3} - 2{m^2} + m - 3{m^2} + 6m - 3 = 0\\
\to \left( {m - 3} \right)\left( {{m^2} - 4m - 3} \right) = 0\\
\to \left[ \begin{array}{l}
m = 3\\
m = 2 + \sqrt 7 \\
m = 2 - \sqrt 7
\end{array} \right. \to \left[ \begin{array}{l}
y = 1\\
y = \frac{{ - 1 + \sqrt 7 }}{3}\\
y = \frac{{ - 1 - \sqrt 7 }}{3}
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\\
x = \frac{{1 + 2\sqrt 7 }}{9}\\
x = \frac{{1 - 2\sqrt 7 }}{9}
\end{array} \right.\\
KL:\left[ \begin{array}{l}
m = 3\\
m = 2 + \sqrt 7 \\
m = 2 - \sqrt 7
\end{array} \right.
\end{array}\)
