a: ĐKXĐ: \(x^3-x^2-4x+4< >0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)< >0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)< >0\)
hay \(x\notin\left\{1;2;-2\right\}\)
b: \(A=\dfrac{2x^2-2}{x^3-x^2-4x+4}\)
\(A=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
Để A=0 thì x+1=0
hay x=-1