Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - 1} \sqrt {\frac{{{x^6} - 7{x^4} + 3}}{{3{x^2} - 7}}} = \sqrt {\frac{{{{\left( 1 \right)}^6} - 7.{{\left( { - 1} \right)}^4} + 3}}{{3.{{\left( { - 1} \right)}^2} - 7}}} = \sqrt {\frac{{ - 3}}{{ - 4}}} = \sqrt {\frac{3}{4}} \Rightarrow b - a = 1\\
\mathop {\lim }\limits_{x \to 3} \frac{{\left( {{x^2} - 9} \right)\sqrt {2{x^2} + 9} }}{{2{x^2} - 5x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x + 3} \right)\sqrt {2{x^2} + 9} }}{{\left( {2x + 1} \right)\left( {x - 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 3} \right)\sqrt {2{x^2} + 9} }}{{2x + 1}} = \frac{{\left( {3 + 3} \right).\sqrt {{{2.3}^2} + 9} }}{{2.3 + 1}} = \frac{{6.3\sqrt 3 }}{7} = \frac{{18\sqrt 3 }}{7}
\end{array}\)
