Cho hàm số \(f\left( x \right) = \left\{ \begin{ar...

D

- Hướng dẫn giải

Ta có $f\left( 9 \right) = 12,\mathop {\lim }\limits_{x \to {9^ - }} {\mkern 1mu} f\left( x \right) = \mathop {\lim }\limits_{x \to {9^ - }} {\mkern 1mu} \left( {ax - 2b} \right) = 9x - 2b,ycbt \Rightarrow \mathop {\lim }\limits_{x \to {9^ - }} {\mkern 1mu} f\left( x \right) = f\left( 9 \right) \Leftrightarrow 9a - 2b = 12$

$\mathop {\lim }\limits_{x \to {9^ - }} {\mkern 1mu} f\left( x \right) = \mathop {\lim }\limits_{x \to {9^ - }} {\mkern 1mu} \frac{{ax - 2b - 12}}{{\sqrt[3]{{x - 1}} - 2}} = \mathop {\lim }\limits_{x \to {9^ - }} {\mkern 1mu} \frac{{\left( {ax - 2b - 12} \right)\left[ {\sqrt[3]{{x - 1}} + 2\sqrt[3]{{x - 1}} + 4} \right]}}{{x - 9}} = 12 \Rightarrow \left\{ \begin{array}{l} a = 1\\  - 2b - 12 =  - 9 \end{array} \right.$

Suy ra $a = 1,b =  - \frac{3}{2}.$. Nên $P = a + b =  - \frac{1}{2}$