a)Tử: \(x^5-2x^4+2x^3-4x^2-3x+6\)
\(=x^5+2x^3-3x-2x^4-4x^2+6\)
\(=x\left(x^4+2x^2-3\right)-2\left(x^4+2x^2-3\right)\)
\(=\left(x-2\right)\left(x^4+2x^2-3\right)\)
\(=\left(x-2\right)\left[x^4-x^2+3x^2-3\right]\)
\(=\left(x-2\right)\left[x^2\left(x^2-1\right)+3\left(x^2-1\right)\right]\)
\(=\left(x-2\right)\left(x^2-1\right)\left(x^2+3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
Mẫu: \(x^2+2x-8=x^2-2x+4x-8\)
\(=x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x+4\right)\)
Suy ra \(A=\dfrac{\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}{\left(x-2\right)\left(x+4\right)}=\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}{x+4}\)
b)\(A=0\Rightarrow\dfrac{\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}{x+4}=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2+3\right)=0\)
Dễ thấy: \(x^2+3\ge3>0\forall x\) (vô nghiệm)
Nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
A có nghĩa khi \(x+4\ne0\Rightarrow x\ne-4\)
A vô nghĩa khi \(x+4=0\Rightarrow x=-4\)
