Biết\(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \dfrac{{a\sqrt 2 }}{b} + c\) với \(a\), \(b\), \(c\)\( \in \mathbb{Z}\) và \(\dfrac{a}{b}\) là phân số tối giản. Giá trị của \(a + b + c\) bằng:

Ta có \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - 2 + 2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - 2}}{{\sqrt 2 \left( {x - 1} \right)}} + \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = I + J\).

Tính \(I = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - 2}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + x + 2 - 4}}{{\sqrt 2 \left( {x - 1} \right)\left( {\sqrt {{x^2} + x + 2}  + 2} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\sqrt 2 \left( {x - 1} \right)\left( {\sqrt {{x^2} + x + 2}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 2}}{{\sqrt 2 \left( {\sqrt {{x^2} + x + 2}  + 2} \right)}} = \dfrac{3}{{4\sqrt 2 }}\).

và \(J = \mathop {\lim }\limits_{x \to 1} \dfrac{{2 - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{8 - 7x - 1}}{{\sqrt 2 \left( {x - 1} \right)\left[ {4 + 2\sqrt[3]{{7x + 1}} + {{\left( {\sqrt[3]{{7x + 1}}} \right)}^2}} \right]}}\)

\(\mathop { = \lim }\limits_{x \to 1} \dfrac{{ - 7}}{{\sqrt 2 \left[ {4 + 2\sqrt[3]{{7x + 1}} + {{\left( {\sqrt[3]{{7x + 1}}} \right)}^2}} \right]}} = \dfrac{{ - 7}}{{12\sqrt 2 }}\).

Do đó \(\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {{x^2} + x + 2}  - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}} = I + J = \dfrac{{\sqrt 2 }}{{12}}\)

Suy ra \(a = 1\), \(b = 12\), \(c = 0\). Vậy \(a + b + c = 13\).