\(A=\dfrac{x^3+3x-4}{x^2+x+4}=\dfrac{x^3+x^2+4x-x^2-x-4}{x^2+x+4}\)
\(A=\dfrac{x\left(x^2+x+4\right)-\left(x^2+x+4\right)}{x^2+x+4}=\dfrac{\left(x-1\right)\left(x^2+x+4\right)}{x^2+x+4}=x-1\)
vậy \(A=x-1\)
\(A=\dfrac{x^3+3x-4}{x^2+x+4}\)
ĐK: \(x^2+x+4\ne0\) đúng với mọi x
\(A=\dfrac{\left(x^3-4\right)+\left(4x-4\right)}{x^2+x+4}=\dfrac{x\left(x-1\right)\left(x+1\right)+4\left(x-1\right)}{x^2+x+4}=x-1\)
KL: \(A=x-1\)