Đáp án:

\[P = 12\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + ax}  - \sqrt[3]{{1 + bx}}}}{x} = 2\\
 \Leftrightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sqrt {1 + ax}  - 1}}{x} + \dfrac{{1 - \sqrt[3]{{1 + bx}}}}{x}} \right] = 2\\
 \Leftrightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\left( {\sqrt {1 + ax}  - 1} \right)\left( {\sqrt {1 + ax}  + 1} \right)}}{{x.\left( {\sqrt {1 + ax}  + 1} \right)}} + \dfrac{{\left( {1 - \sqrt[3]{{1 + bx}}} \right)\left( {{1^2} + 1.\sqrt[3]{{1 + bx}} + {{\sqrt[3]{{1 + bx}}}^2}} \right)}}{{{1^2} + 1.\sqrt[3]{{1 + bx}} + {{\sqrt[3]{{1 + bx}}}^2}}}} \right] = 2\\
 \Leftrightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\left( {1 + ax} \right) - {1^2}}}{{x.\left( {\sqrt {1 + ax}  + 1} \right)}} + \dfrac{{{1^3} - \left( {1 + bx} \right)}}{{1 + \sqrt[3]{{1 + bx}} + {{\sqrt[3]{{1 + bx}}}^2}}}} \right] = 2\\
 \Leftrightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{ax}}{{x\left( {\sqrt {1 + ax}  + 1} \right)}} + \dfrac{{ - bx}}{{1 + \sqrt[3]{{1 + bx}} + {{\sqrt[3]{{1 + bx}}}^2}}}} \right] = 2\\
 \Leftrightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{a}{{\sqrt {1 + ax}  + 1}} - \dfrac{b}{{1 + \sqrt[3]{{1 + bx}} + {{\sqrt[3]{{1 + bx}}}^2}}}} \right] = 2\\
 \Leftrightarrow \dfrac{a}{{\sqrt 1  + 1}} - \dfrac{b}{{1 + 1 + 1}} = 2\\
 \Leftrightarrow \dfrac{a}{2} - \dfrac{b}{3} = 2\\
 \Leftrightarrow \dfrac{{3a - 2b}}{6} = 2\\
 \Leftrightarrow 3a - 2b = 12
\end{array}\)