a) Ta có: \(A=\left(\frac{1}{1-x}+\frac{x}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{-1}{x-1}+\frac{x}{x+1}+\frac{5-x}{x^2-1}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{-1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x-1+x^2-x+5-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{x^2-3x+4}{1-2x}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\notin\left\{1;-1\right\}\\x\ne\frac{1}{2}\end{matrix}\right.\)
Để |A|=A thì \(A\ge0\)
\(\Leftrightarrow\frac{x^2-3x+4}{1-2x}\ge0\)
mà \(\frac{x^2-3x+4}{1-2x}\ne0\forall x\) thỏa mãn ĐKXĐ
nên \(\frac{x^2-3x+4}{1-2x}>0\)
\(\Leftrightarrow1-2x>0\)(Vì \(x^2-3x+4>0\forall x\))
\(\Leftrightarrow-2x>-1\)
hay \(x< \frac{1}{2}\)(nhận)
Vậy: Khi \(x< \frac{1}{2}\) thì |A|=A
