Đáp án:
$\dfrac{x^2+5}{x^3-3x-2}=\dfrac{1}{x-2}-\dfrac{2}{(x+1)^2}$
Giải thích các bước giải:
$\begin{split}\dfrac{x^2+5}{x^3-3x-2}&=\dfrac{x^2+5}{x^3-2^3-(3x-6)}\\&=\dfrac{x^2+5}{(x-2)(x^2+2x+4)-3(x-2)}\\&=\dfrac{x^2+5}{(x-2)(x^2+2x+1)}\\&=\dfrac{x^2+2x+1-2(x-2)}{(x-2)(x+1)^2}\\&=\dfrac{(x+1)^2-2(x-2)}{(x-2)(x+1)^2}\\&=\dfrac{(x+1)^2}{(x-2)(x+1)^2}-\dfrac{2(x-2)}{(x-2)(x+1)^2}\\&=\dfrac{1}{x-2}-\dfrac{2}{(x+1)^2}\end{split}$
$\rightarrow a=1,b=-2$