Câu 13:
limx→4−−x2−3x−1∣x−4∣\mathop {\lim }\limits_{x \to {4^ - }} \dfrac{{ - {x^2} - 3x - 1}}{{\left| {x - 4} \right|}}x→4−lim∣x−4∣−x2−3x−1 bằng:
