Câu hỏi 55

Nhân liên hợp để khử dạng  \(\frac{0}{0}\)  rồi tính giới hạn của biểu thức.

\(\begin{array}{l}\frac{{\sqrt[3]{{x + 7}} - \sqrt {{x^2} + x + 2} }}{{x - 1}} = \frac{{\sqrt[3]{{x + 7}} - 2}}{{x - 1}} - \frac{{\sqrt[{}]{{{x^2} + x + 2}} - 2}}{{x - 1}}\\ = \frac{{\left( {\sqrt[3]{{x + 7}} - 2} \right)\left( {\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}}{{\left( {x - 1} \right)\left( {\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}} - \frac{{\left( {\sqrt[{}]{{{x^2} + x + 2}} - 2} \right)\left( {\sqrt[{}]{{{x^2} + x + 2}} + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt[{}]{{{x^2} + x + 2}} + 2} \right)}}\\ = \frac{{x + 7 - 8}}{{\left( {x - 1} \right)\left( {\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}} - \frac{{{x^2} + x + 2 - 4}}{{\left( {x - 1} \right)\left( {\sqrt[{}]{{{x^2} + x + 2}} + 2} \right)}}\\ = \frac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}} - \frac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt[{}]{{{x^2} + x + 2}} + 2} \right)}}\\ = \frac{1}{{\sqrt[3]{{{{\left( {x + 7} \right)}^2}}} + 2\sqrt[3]{{\left( {x + 7} \right)}} + 4}} - \frac{{x + 2}}{{\sqrt[{}]{{{x^2} + x + 2}} + 2}}\end{array}\)