Phương pháp giải:

Thêm bớt x, nhân liên hợp.

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - \sqrt[3]{{{x^3} - {x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - x + x - \sqrt[3]{{{x^3} - {x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + x}  - x} \right) + \mathop {\lim }\limits_{x \to  + \infty } \left( {x - \sqrt[3]{{{x^3} - {x^2}}}} \right)\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2} + x - {x^2}}}{{\sqrt {{x^2} + x}  + x}} + \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^3} - {x^3} + {x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} - {x^2}}} + {{\sqrt[3]{{{x^3} - {x^2}}}}^2}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{x}{{\sqrt {{x^2} + x}  + x}} + \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^2}}}{{{x^2} + x\sqrt[3]{{{x^3} - {x^2}}} + {{\sqrt[3]{{{x^3} - {x^2}}}}^2}}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \dfrac{1}{{\sqrt {1 + \dfrac{1}{x}}  + 1}} + \mathop {\lim }\limits_{x \to  + \infty } \dfrac{1}{{1 + \sqrt[3]{{1 - \dfrac{1}{x}}} + {{\sqrt[3]{{1 - \dfrac{1}{x}}}}^2}}}\\ = \dfrac{1}{{1 + 1}} + \dfrac{1}{{1 + 1 + 1}} = \dfrac{5}{6}.\end{array}\)

Chọn D.