Giới hạn \(\mathop {\lim }\limits_{x \to 2} \left(...

D

- Hướng dẫn giải

Phương pháp giải:

Tính \(\mathop {\lim }\limits_{x \to 2} \left( {\frac{1}{{3{x^2} - 4x - 4}} + \frac{1}{{{x^2} - 12x + 20}}} \right)\) bằng cách phân tích:

\(\begin{array}{l}\frac{1}{{3{x^2} - 4x - 4}} + \frac{1}{{{x^2} - 12x + 20}} = \frac{1}{{\left( {x - 2} \right)\left( {3x + 2} \right)}} + \frac{1}{{\left( {x - 2} \right)\left( {x - 10} \right)}}\\ = \frac{{x - 10 + 3x + 2}}{{\left( {x - 2} \right)\left( {3x + 2} \right)\left( {x - 10} \right)}} = \frac{{4\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3x + 2} \right)\left( {x - 10} \right)}} = \frac{4}{{\left( {3x + 2} \right)\left( {x - 10} \right)}}.\end{array}\)

Giải chi tiết:

Ta có :  \(\frac{1}{{3{x^2} - 4x - 4}} + \frac{1}{{{x^2} - 12x + 20}} = \frac{1}{{\left( {x - 1} \right)\left( {3x + 2} \right)}} + \frac{1}{{\left( {x - 2} \right)\left( {x - 10} \right)}}\)

            \( = \frac{{x - 10 + 3x + 2}}{{\left( {x - 2} \right)\left( {3x + 2} \right)\left( {x - 10} \right)}} = \frac{{4\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3x + 2} \right)\left( {x - 10} \right)}} = \frac{4}{{\left( {3x + 2} \right)\left( {x - 10} \right)}}\)

Do đó:  \(\mathop {\lim }\limits_{x \to 2} \left( {\frac{1}{{3{x^2} - 4x - 4}} + \frac{1}{{{x^2} - 12x + 20}}} \right) = \mathop {\lim }\limits_{x \to 2} \frac{4}{{\left( {3x + 2} \right)\left( {x - 10} \right)}} = \frac{4}{{\left( {3.2 + 2} \right)\left( {2 - 10} \right)}} = \frac{{ - 1}}{{16}}.\)

Vậy theo bài ra thì \(a =  - 1,\,\,b = 16\) nên  \(b - a = 17.\)

Chọn D.