ĐKXĐ: x khác 0; x khác -2;2;4
A=\(\dfrac{x^3+x^2-6x}{x^3-4x}=\dfrac{x^2+x-6}{x^2-4}\)
\(=\dfrac{x^2-2x+3x-6}{\left(x-2\right)\left(x+2\right)}\)\(=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+3}{x+2}\)
Với x=98, ta có:
A\(=\dfrac{x+3}{x+2}=\dfrac{98+3}{98+2}=\dfrac{101}{100}\)