\(A=\left(\frac{\left(1-x\right)\left(x-1\right)-\left(x+3\right)^2}{\left(x+3\right)\left(x-1\right)}\right):\left(\frac{\left(x+3\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+3\right)}\right)\)(ĐKXĐ: x khác -3 và 1)
\(=\frac{-x^2+2x-1-x^2-6x-9}{\left(x+3\right)\left(x-1\right)}:\frac{x^2+6x+9-x^2+2x-1}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{-2x^2-4x-10}{\left(x+3\right)\left(x-1\right)}:\frac{8x+8}{\left(x-1\right)\left(x+3\right)}\)
\(=\frac{-2x^2-4x-10}{8x+8}\)
Mà \(-2x^2-4x-10=-2\left(x+1\right)^2-8< 0\forall x\)
Nên để A < 0 thì \(8x+8>0\Leftrightarrow x>-1\)
Vậy với \(x>-1,x\ne1\)thì A < 0