Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}12{\rm{     }}\left( {x \ge 9} \right)\\\frac{{ax - 2b - 12}}{{\sqrt[3]{{

* 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 \(P = a + b =  - \frac{1}{2}\)