\(1)\left\{ \begin{array}{l}\frac{1}{{\sqrt {x - 1} }} - \frac{1} = 3\\\frac{3}{{\sqrt {x - 1} }} + \frac{2} = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sqrt {x - 1} = 1\\y + 2 = - \frac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 2\\y = - \frac{5}{2}\end{array} \right.\)

\({x^2} - 2x - 3 = 0 \Leftrightarrow \left[ \begin{array}{l}{x_A} = 3 \Rightarrow A\left( {3;9} \right)\\{x_B} = - 1 \Rightarrow B\left( { - 1;1} \right)\end{array} \right.\)

b) \(C\left( {x;{x^2}} \right)\)và \(A',B',C'\)lần lượt là chân đường cao hạ xuống \(Ox\)

\(\begin{array}{l}{S_{ABC}} = {S_{AA'B'B}} + {S_{ACC'A'}} - {S_{BCC'B'}} = 2{x^2} + 4x + 6 = 8 - 2{\left( {x - 1} \right)^2} \le 8\\ \Rightarrow MaxS = 8 \Leftrightarrow C\left( {1;1} \right)\end{array}\)