Giải thích các bước giải:
$\lim_{x\to-\dfrac12}\dfrac{-3x^2+ax+b}{x-1}=-4$
$\to -3x^2+ax+b=0$ có nghiệm $x=1$
$\to -3+a+b=0\to a+b=3$
$\to\lim_{x\to-\dfrac12}\dfrac{-3x^2+ax+3-a}{x-1}=-4$
$\to\lim_{x\to-\dfrac12}\dfrac{-3(x^2-1)+a(x-1)}{x-1}=-4$
$\to\lim_{x\to-\dfrac12}\dfrac{-3(x-1)(x+1)+a(x-1)}{x-1}=-4$
$\to\lim_{x\to-\dfrac12}-3(x+1)+a=-4$
$\to-3(-\dfrac{1}{2}+1)+a=-4$
$\to a=-\dfrac52\to b=3-a=\dfrac{11}{2}$
$\to a-b=8$
