Cho hàm số sau (y = dfrac{{2x – 1}}{{2x – 2}}) có đồ thị (left( C right)).

(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{{ – 2left( {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 Aleft( {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 2left( {x – {x_0}} right) = left( {2{x_0} – 1} right)left( {2{x_0} – 2} right) – {left( {2{x_0} – 2} right)^2}\ Leftrightarrow 2left( {x – {x_0}} right) = 2{x_0} – 2 Leftrightarrow x = 2{x_0} – 1\ Rightarrow Bleft( {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 (Ileft( {1;1} right))

Ta có: ({S_{Delta OIB}} = 8{S_{Delta OIA}} Leftrightarrow dfrac{1}{2}.dleft( {B;OI} right).OI = 8.dfrac{1}{2}.dleft( {B;OI} right).OI Leftrightarrow dleft( {B;OI} right) = 8.dleft( {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| = 8left| {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 =  – 2end{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