Đáp án:

\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{x + 7}} - \sqrt {x + 3} }}{{{x^2} - 3x + 2}} = \frac{1}{6}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{x + 7}} - \sqrt {x + 3} }}{{{x^2} - 3x + 2}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sqrt[3]{{x + 7}} - 2}}{{{x^2} - 3x + 2}} + \frac{{2 - \sqrt {x + 3} }}{{{x^2} - 3x + 2}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {\sqrt[3]{{x + 7}} - 2} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2.\sqrt[3]{{x + 7}} + {2^2}} \right)}}{{\left( {{x^2} - 3x + 2} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2.\sqrt[3]{{x + 7}} + 4} \right)}} + \frac{{\left( {2 - \sqrt {x + 3} } \right)\left( {2 + \sqrt {x + 3} } \right)}}{{\left( {{x^2} - 3x + 2} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {x + 7} \right) - {2^3}}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4} \right)}} + \frac{{{2^2} - \left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{x - 1}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4} \right)}} + \frac{{1 - x}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{1}{{\left( {x - 2} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4} \right)}} + \frac{{ - 1}}{{\left( {x - 2} \right)\left( {2 + \sqrt {x + 3} } \right)}}} \right]\\
 = \frac{1}{{\left( {1 - 2} \right)\left( {{{\sqrt[3]{{1 + 7}}}^2} + 2\sqrt[3]{{1 + 7}} + 4} \right)}} - \frac{1}{{\left( {1 - 2} \right)\left( {2 + \sqrt {1 + 3} } \right)}}\\
 = \frac{1}{6}
\end{array}\)