\(y = \dfrac{{2x - 1}}{{2x - 2}} \Rightarrow y' = \dfrac{{ - 2}}{{{{\left( {2x - 2} \right)}^2}}}\)

Phương trình tiếp tuyến của (C) tại M là: \(y = \dfrac{{ - 2}}{{{{\left( {2{x_0} - 2} \right)}^2}}}.\left( {x - {x_0}} \right) + \dfrac{{2{x_0} - 1}}{{2{x_0} - 2}}\)

Cho \(x = 1\)

\(\begin{array}{l} \Rightarrow y = \dfrac{{ - 2}}{{{{\left( {2{x_0} - 2} \right)}^2}}}.\left( {1 - {x_0}} \right) + \dfrac{{2{x_0} - 1}}{{2{x_0} - 2}} = \dfrac{{ - 2\left( {1 - {x_0}} \right)}}{{{{\left( {2{x_0} - 2} \right)}^2}}} + \dfrac{{\left( {2{x_0} - 1} \right)\left( {2{x_0} - 2} \right)}}{{2{x_0} - 2}}\\ = \dfrac{{4x_0^2 - 4{x_0}}}{{{{\left( {2{x_0} - 2} \right)}^2}}} = \dfrac{{{x_0}}}{{{x_0} - 1}} \Rightarrow A\left( {1;\dfrac{{{x_0}}}{{{x_0} - 1}}} \right)\end{array}\)

Cho \(y = 1\)

\(\begin{array}{l} \Rightarrow 1 = \dfrac{{ - 2}}{{{{\left( {2{x_0} - 2} \right)}^2}}}.\left( {x - {x_0}} \right) + \dfrac{{2{x_0} - 1}}{{2{x_0} - 2}}\\ \Leftrightarrow 2\left( {x - {x_0}} \right) = \left( {2{x_0} - 1} \right)\left( {2{x_0} - 2} \right) - {\left( {2{x_0} - 2} \right)^2}\\ \Leftrightarrow 2\left( {x - {x_0}} \right) = 2{x_0} - 2 \Leftrightarrow x = 2{x_0} - 1\\ \Rightarrow B\left( {2{x_0} - 1;1} \right)\end{array}\)

Đồ thị \(\left( C \right)\) có TCĐ là \(x = 1\) và TCN là \(y = 1\), giao điểm của 2 đường tiệm cận \(I\left( {1;1} \right)\)

Ta có: \({S_{\Delta OIB}} = 8{S_{\Delta OIA}} \Leftrightarrow \dfrac{1}{2}.d\left( {B;OI} \right).OI = 8.\dfrac{1}{2}.d\left( {B;OI} \right).OI \Leftrightarrow d\left( {B;OI} \right) = 8.d\left( {B;OI} \right)\) (*)

Phương trình đường thẳng OI là: \(y = x \Leftrightarrow x - y = 0\)

 (*)\( \Leftrightarrow \dfrac{{\left| {2{x_0} - 1 - 1} \right|}}{{\sqrt 2 }} = 8.\dfrac{{\left| {1 - \dfrac{{{x_0}}}{{{x_0} - 1}}} \right|}}{{\sqrt 2 }} \Leftrightarrow \left| {2{x_0} - 2} \right| = 8\left| {\dfrac{{ - 1}}{{{x_0} - 1}}} \right| \Leftrightarrow {\left( {{x_0} - 1} \right)^2} = 4\)\( \Leftrightarrow \left[ \begin{array}{l}{x_0} - 1 = 2\\{x_0} - 1 =  - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0} = 3\\{x_0} =  - 1(L)\end{array} \right.\)

\( \Rightarrow {y_0} = \dfrac{{2.3 - 1}}{{2.3 - 2}} = \dfrac{5}{4}\)\( \Rightarrow S = {x_0} + 4{y_0} = 8\).

Chọn: A