Đáp án:
\[P = 3\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \mathop {\lim }\limits_{x \to 1} \frac{{2{x^2} + \left( {m - 2} \right)x - m}}{{{x^2} - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {2{x^2} - 2x} \right) + \left( {mx - m} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{2x\left( {x - 1} \right) + m\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {2x + m} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{2x + m}}{{x + 1}}\\
= \frac{{2.1 + m}}{{1 + 1}} = \frac{{m + 2}}{2}\\
A = 2 \Leftrightarrow \frac{{m + 2}}{2} = 2 \Leftrightarrow m = 2\\
P = \frac{{{m^3} + 1}}{{m + 1}} = \frac{{{2^3} + 1}}{{2 + 1}} = 3
\end{array}\)
