Bài 125:
a) Tọa độ giao điểm là nghiệm của hệ phương trình:
\(\begin{array}{l}\left\{ \begin{array}{l}\left( {a + 1} \right)x - 2y - a - 1 = 0\\x + \left( {a - 1} \right)y - {a^2} = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x + \left( {a - 1} \right)y - {a^2} = 0\\\left( {a + 1} \right)x + \left( {{a^2} - 1} \right)y - {a^2}\left( {a + 1} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x + \left( {a - 1} \right)y - {a^2} = 0\\\left( {{a^2} - 1 + 2} \right)y - {a^2}\left( {a + 1} \right) + a + 1 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x + \left( {a - 1} \right)y - {a^2} = 0\\\left( {{a^2} + 1} \right)y - \left( {{a^2} - 1} \right)\left( {a + 1} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x + \left( {a - 1} \right)y - {a^2} = 0\\y = \dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = {a^2} - \left( {a - 1} \right).\dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}}\\y = \dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{{a^2}\left( {{a^2} + 1} \right) - {{\left( {{a^2} - 1} \right)}^2}}}{{{a^2} + 1}}\\y = \dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = \dfrac{{3{a^2} - 1}}{{{a^2} + 1}}\\y = \dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}}\end{array} \right.\\ \Rightarrow I\left( {\dfrac{{3{a^2} - 1}}{{{a^2} + 1}};\dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}}} \right)\end{array}\)
b) Đường thẳng MN: \(\dfrac{x}{a} + \dfrac{y}{a} = 1 \Leftrightarrow x + y - a = 0\) với \(a \ne 0\)
\(I \in MN \Leftrightarrow \dfrac{{3{a^2} - 1}}{{{a^2} + 1}} + \dfrac{{\left( {{a^2} - 1} \right)\left( {a + 1} \right)}}{{{a^2} + 1}} - a = 0\)
\(\begin{array}{l} \Leftrightarrow 3{a^2} - 1 + {a^3} + {a^2} - a - 1 - {a^3} - a = 0\\ \Leftrightarrow 4{a^2} - 2a - 2 = 0\\ \Leftrightarrow \left[ \begin{array}{l}a = 1\\a = - \dfrac{1}{2}\end{array} \right.\end{array}\)
Vậy \(a = 1\) hoặc \(a = - \dfrac{1}{2}\)
