Đáp án đúng: D

Phương pháp giải:

- Tính \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\).- Sử dụng phương pháp nhân liên hợp.Giải chi tiết:Đặt \(g\left( x \right) = \dfrac{{f\left( x \right) - 20}}{{x - 2}}\) ta có \(\mathop {\lim }\limits_{x \to 2} g\left( x \right) = 10\) và \(f\left( x \right) - 20 = g\left( x \right)\left( {x - 2} \right) \Leftrightarrow f\left( x \right) = g\left( x \right)\left( {x - 2} \right) + 20\).\(\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {g\left( x \right)\left( {x - 2} \right) + 20} \right] = 10.\left( {2 - 2} \right) + 20 = 20\).Ta có:\(\begin{array}{l}\,\,\,\,\,\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{6f\left( x \right) + 5 - 125}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{6\left[ {f\left( x \right) - 20} \right]}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{f\left( x \right) - 20}}{{x - 2}}\dfrac{6}{{\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\\ = 10.\dfrac{6}{{\left( {2 + 3} \right)\left[ {{{\left( {\sqrt[3]{{6.20 + 5}}} \right)}^2} + 5\sqrt[3]{{6.20 + 5}} + 25} \right]}} = \dfrac{4}{{25}}\end{array}\)Chọn D