Cho đa thức \(f\left( x \right)\) thoả mãn \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 20}}{{x - 2}} = 10.\) Tính \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}}\).

Ta có \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 20}}{{x - 2}} = 10 \Rightarrow f\left( 2 \right) - 20 = 0 \Leftrightarrow f\left( 2 \right) = 20\).

Lại có \(\frac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}} = \frac{{6f\left( x \right) + 5 - 125}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\sqrt[3]{{6f\left( x \right) + 5}}}^2} + \sqrt[3]{{6f\left( x \right) + 5}} \cdot 5 + {5^2}} \right]}}\)

\[ = \frac{{6\left[ {f\left( x \right) + 20} \right]}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {\sqrt[3]{{6f\left( x \right) + {5^2}}} + \sqrt[3]{{6f(x) + 5}}.5 + {5^2}} \right]}}\]

\( = 6 \cdot \frac{{f\left( x \right) + 20}}{{x - 2}} \cdot \frac{1}{{\left( {x + 3} \right)\left[ {{{\sqrt[3]{{6f\left( x \right) + 5}}}^2} + \sqrt[3]{{6f\left( x \right) + 5}} \cdot 5 + {5^2}} \right]}}\)

Suy ra \(\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{6f(x) + 5}} - 5}}{{{x^2} + x - 6}} = \mathop {\lim }\limits_{x \to 2} \left( {6.\frac{{f\left( x \right) + 20}}{{x - 2}} \cdot \frac{1}{{(x + 3)\left[ {\sqrt[3]{{6f\left( x \right) + 5}} + \sqrt[3]{{6f\left( x \right) + 5}} \cdot 5 + {5^2}} \right]}}} \right)\)

=6. \(\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 20}}{{x - 2}}.\mathop {\lim }\limits_{x \to 2} \frac{1}{{\left( {x + 3} \right)\left[ {\sqrt[3]{{6f\left( x \right) + 5}} + \sqrt[3]{{6f\left( x \right) + 5}} \cdot 5 + {5^2}} \right]}}\)

Điền đáp án \(\frac{4}{{25}}.\)