Đáp án đúng: B

Phương pháp giải:
Sử dụng giới hạn đặc biệt \(\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\).
Giải chi tiết:Ta có:
\(\begin{array}{l}\,\,\,\dfrac{{1 - \sqrt {\cos x.cos2x.cos3x} }}{{{x^2}}}\\ = \dfrac{{1 - \cos x.cos2x.cos3x}}{{{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{1 - \dfrac{1}{2}cos2x.\left( {\cos 4x + \cos 2x} \right)}}{{{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{2 - cos2x.\cos 4x - {{\cos }^2}2x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{2 - \dfrac{1}{2}\left( {\cos 6x + \cos 2x} \right) - {{\cos }^2}2x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{{{\sin }^2}2x + 1 - \dfrac{1}{2}\left( {\cos 6x + \cos 2x} \right)}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{{{\sin }^2}2x + 1 - \dfrac{1}{2}\cos 6x - \dfrac{1}{2} + {{\sin }^2}x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{{{\sin }^2}2x + {{\sin }^2}x + \dfrac{1}{2} - \dfrac{1}{2}\cos 6x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{{{\sin }^2}2x + {{\sin }^2}x + \dfrac{1}{2}.2{{\sin }^2}3x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ = \dfrac{{{{\sin }^2}2x + {{\sin }^2}x + {{\sin }^2}3x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\end{array}\)
\(\begin{array}{l}\,\,\,\,\,\,L = \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \sqrt {\cos x.cos2x.cos3x} }}{{{x^2}}}\\ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x + {{\sin }^2}x + {{\sin }^2}3x}}{{2{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ \Rightarrow L = \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}} + \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\\,\,\,\,\,\,\,\,\,\,\, + \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}3x}}{{{x^2}\left( {1 + \sqrt {\cos x.cos2x.cos3x} } \right)}}\\ \Rightarrow L = \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}2x}}{{{{\left( {2x} \right)}^2}}}.\dfrac{4}{{1 + \sqrt {\cos x.cos2x.cos3x} }} + \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}x}}{{{x^2}}}.\dfrac{1}{{1 + \sqrt {\cos x.cos2x.cos3x} }}\\\,\,\,\,\,\,\,\,\,\,\,\, + \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sin }^2}3x}}{{{{\left( {3x} \right)}^2}}}.\dfrac{9}{{1 + \sqrt {\cos x.cos2x.cos3x} }}\\ \Rightarrow L = \dfrac{1}{2}.\dfrac{4}{{1 + 1}} + \dfrac{1}{2}.\dfrac{1}{{1 + 1}} + \dfrac{1}{2}.\dfrac{9}{{1 + 1}} = \dfrac{7}{2}\end{array}\)
Vậy \(L = \dfrac{7}{2}\).