Cho \(a,\,\,b\) là các số nguyên và \(\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}} = 20\). Tính \(P = {a^2} + {b^2} - a - b\).

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \dfrac{{a\left( {{x^2} - 1} \right) + b\left( {x - 1} \right) + a + b - 5}}{{x - 1}}\\ = \mathop {\lim }\limits_{x \to 1} \left[ {a\left( {x + 1} \right) + b + \dfrac{{a + b - 5}}{{x - 1}}} \right]\\ = \mathop {\lim }\limits_{x \to 1} \left[ {a\left( {x + 1} \right) + b} \right] + \mathop {\lim }\limits_{x \to 1} \dfrac{{a + b - 5}}{{x - 1}}\\ = 2a + b + \mathop {\lim }\limits_{x \to 1} \dfrac{{a + b - 5}}{{x - 1}}\end{array}\)

Theo bài ra ta có:

\(\mathop {\lim }\limits_{x \to 1} \dfrac{{a{x^2} + bx - 5}}{{x - 1}} = 20 \Leftrightarrow \left\{ \begin{array}{l}2a + b = 20\\a + b - 5 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 15\\b =  - 10\end{array} \right.\).

Vậy \(P = {a^2} + {b^2} - a - b = {15^2} + {\left( { - 10} \right)^2} - 15 - \left( { - 10} \right) = 320\).

Chọn D.