Câu hỏi 9

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {2x + 3}  - 3}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {\sqrt {2x + 3}  - 3} \right)\left( {\sqrt {2x + 3}  + 3} \right)}}{{\left( {x - 3} \right)\left( {\sqrt {2x + 3}  + 3} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{2x - 6}}{{\left( {x - 3} \right)\left( {\sqrt {2x + 3}  + 3} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{2}{{\sqrt {2x + 3}  + 3}} = \frac{2}{{\sqrt {2.3 + 3}  + 3}} = \frac{1}{3}\\ \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 3\end{array} \right. \Rightarrow P = {a^2} + {b^2} = {1^2} + {3^2} = 10\end{array}\)

Chọn: A.