Tính a + b.
Giải thích
\(\mathop {\lim }\limits_{x \to - \sqrt 3 } \frac{{2{x^3} + 6\sqrt 3 }}{{3 - {x^2}}} = \mathop {\lim }\limits_{x \to - \sqrt 3 } \frac{{2\left( {{x^3} + {{\left( {\sqrt 3 } \right)}^3}} \right)}}{{3 - {x^2}}} = \mathop {\lim }\limits_{x \to - \sqrt 3 } \frac{{2\left( {x + \sqrt 3 } \right)\left( {{x^2} - \sqrt 3 x + 3} \right)}}{{\left( {\sqrt 3 - x} \right)\left( {\sqrt 3 + x} \right)}}\)
\( = \mathop {\lim }\limits_{x \to - \sqrt 3 } \frac{{2\left( {{x^2} - \sqrt 3 x + 3} \right)}}{{\sqrt 3 - x}}\)\( = \frac{{2\left( {3 + 3 + 3} \right)}}{{2\sqrt 3 }} = \frac{9}{{\sqrt 3 }} = 3\sqrt 3 \).
Suy ra a = 3; b = 0. Do đó a + b = 3.
Trả lời: 3.