Lời giải chi tiết:

ĐKXĐ: \(x \in \mathbb{R}\backslash \left\{ { - \dfrac{3}{2}} \right\}\). Ta có: \(y' = \dfrac{{ - 1}}{{{{\left( {2x + 3} \right)}^2}}}\).

Khi đó phương trình tiếp tuyến của đồ thị \(\left( C \right)\) tại điểm có hoành độ \(x = {x_0}\) là:

\(y = \dfrac{{ - 1}}{{{{\left( {2{x_0} + 3} \right)}^2}}}\left( {x - {x_0}} \right) + \dfrac{{{x_0} + 2}}{{2{x_0} + 3}}\)

Cho \(y = 0 \Rightarrow 0 = \dfrac{{ - 1}}{{{{\left( {2{x_0} + 3} \right)}^2}}}\left( {x - {x_0}} \right) + \dfrac{{{x_0} + 2}}{{2{x_0} + 3}}\)

\(\begin{array}{l}y = 0 \Rightarrow 0 = \dfrac{{ - 1}}{{{{\left( {2{x_0} + 3} \right)}^2}}}\left( {x - {x_0}} \right) + \dfrac{{{x_0} + 2}}{{2{x_0} + 3}}\\ \Leftrightarrow  - \left( {x - {x_0}} \right) + \left( {{x_0} + 2} \right)\left( {2{x_0} + 3} \right) = 0\\ \Leftrightarrow x = 2{x_0}^2 + 8{x_0} + 6\\ \Rightarrow A\left( {2{x_0}^2 + 8{x_0} + 6;0} \right) \Rightarrow OA = \left| {2{x_0}^2 + 8{x_0} + 6} \right|\end{array}\)

Cho \(x = 0 \Rightarrow y = \dfrac{{{x_0}}}{{{{\left( {2{x_0} + 3} \right)}^2}}} + \dfrac{{{x_0} + 2}}{{2{x_0} + 3}} = \dfrac{{2{x_0}^2 + 8{x_0} + 6}}{{{{(2{x_0} + 3)}^2}}}\)

\(\begin{array}{l} \Rightarrow B\left( {0;\dfrac{{2{x_0}^2 + 8{x_0} + 6}}{{{{\left( {2{x_0} + 3} \right)}^2}}}} \right)\\ \Rightarrow OB = \left| {\dfrac{{2{x_0}^2 + 8{x_0} + 6}}{{{{\left( {2{x_0} + 3} \right)}^2}}}} \right| = \dfrac{{\left| {2{x_0}^2 + 8{x_0} + 6} \right|}}{{{{\left( {2{x_0} + 3} \right)}^2}}}\end{array}\)  

Theo bài ra ta có: \(\Delta OAB\) cân tại \(O\) nên \(OA = OB\)

\(\begin{array}{l} \Leftrightarrow \left| {2{x_0}^2 + 8{x_0} + 6} \right| = \dfrac{{\left| {2{x_0}^2 + 8{x_0} + 6} \right|}}{{{{\left( {2{x_0} + 3} \right)}^2}}}\\ \Leftrightarrow \left| {2{x_0}^2 + 8{x_0} + 6} \right|\left( {1 - \dfrac{1}{{{{\left( {2{x_0} + 3} \right)}^2}}}} \right) = 0\\ \Leftrightarrow {\left( {2{x_0} + 3} \right)^2} = 1\\ \Leftrightarrow \left[ \begin{array}{l}2{x_0} + 3 = 1\\2{x_0} + 3 =  - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}{x_0} =  - 1\\{x_0} =  - 2\end{array} \right.\,\,\,\left( {tm} \right)\end{array}\)

Ta không xét \(\left| {2x_0^2 + 8{x_0} + 6} \right| = 0\) vì khi đó \(OA = OB = 0\) hay \(O \equiv A \equiv B\).

Khi đó ta có \({M_1}\left( { - 1;1} \right),\,\,{M_2}\left( { - 2;0} \right)\).

Vậy \({M_1}{M_2} = \sqrt {{{\left( { - 2 + 1} \right)}^2} + {{\left( {0 - 1} \right)}^2}}  = \sqrt 2 \).

Chọn C.