Đáp án:
\[a + b + c = 13\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {{x^2} + x + 2} - \sqrt[3]{{7x + 1}}}}{{\sqrt 2 \left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {{x^2} + x + 2} - 2} \right) + \left( {2 - \sqrt[3]{{7x + 1}}} \right)}}{{\sqrt 2 \left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{{x^2} + x + 2 - {2^2}}}{{\sqrt {{x^2} + x + 2} + 2}} + \frac{{{2^3} - \left( {7x + 1} \right)}}{{4 + 2.\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}}}{{\sqrt 2 \left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\sqrt {{x^2} + x + 2} + 2}} - \frac{{7\left( {x - 1} \right)}}{{4 + 2\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}}}{{\sqrt 2 \left( {x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt 2 }}.\left( {\frac{{x + 2}}{{\sqrt {{x^2} + x + 2} + 2}} - \frac{7}{{4 + 2.\sqrt[3]{{7x + 1}} + {{\sqrt[3]{{7x + 1}}}^2}}}} \right)\\
= \frac{1}{{\sqrt 2 }}.\left( {\frac{{1 + 2}}{{\sqrt {{1^2} + 1 + 2} + 2}} - \frac{7}{{4 + 2.\sqrt[3]{{7.1 + 1}} + {{\sqrt[3]{{7.1 + 1}}}^2}}}} \right)\\
= \frac{1}{{\sqrt 2 }}.\left( {\frac{3}{4} - \frac{7}{{12}}} \right)\\
= \frac{{\sqrt 2 }}{{12}}\\
\Rightarrow a = 1;\,\,\,b = 12;\,\,\,c = 0\\
\Rightarrow a + b + c = 13
\end{array}\)
