ĐKXĐ: \(x\ne\left\{\pm1;\pm3\right\}\)
\(A=\left(\frac{\left(1-x\right)\left(x-1\right)-\left(x+3\right)^2}{\left(x+3\right)\left(x-1\right)}\right):\left(\frac{\left(x+3\right)^2-\left(x-1\right)^2}{\left(x+3\right)\left(x-1\right)}\right)\)
\(A=\frac{\left(-2x^2-4x-10\right)}{\left(x+3\right)\left(x-1\right)}:\frac{\left(8x+8\right)}{\left(x+3\right)\left(x-1\right)}\)
\(A=\frac{-2\left(x^2+2x+5\right)}{\left(x+3\right)\left(x-1\right)}.\frac{\left(x+3\right)\left(x-1\right)}{8\left(x+1\right)}=\frac{-\left(x^2+2x+5\right)}{4\left(x+1\right)}\)
Do \(-\left(x^2+2x+5\right)=-\left(x+1\right)^2-4< 0\) \(\forall x\)
\(\Rightarrow A< 0\Leftrightarrow x+1>0\Rightarrow x>-1\)
Vậy để A nhận giá trị âm thì \(\left\{{}\begin{matrix}x>-1\\x\ne\left\{1;3\right\}\end{matrix}\right.\)