a) ĐKXĐ: x∉{1;-1}
b) Ta có: \(B=\frac{x-1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{x^2-1}\)
\(=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-1+x^2-3x+2-2x^2-x-5}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-4x-4}{\left(x-1\right)\left(x+1\right)}=\frac{-4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{-4}{x-1}\)
c) Để B<2 thì \(\frac{-4}{x-1}< 2\)
\(\Leftrightarrow\frac{-4}{x-1}< \frac{2\left(x-1\right)}{x-1}\)
\(\Leftrightarrow-4< 2\left(x-1\right)\)
\(\Leftrightarrow-4< 2x-2\)
\(\Leftrightarrow-4-2x+2< 0\)
\(\Leftrightarrow-2x-2< 0\)
\(\Leftrightarrow-2x< 2\)
hay x>-1
Vậy: Để B<2 thì S={x>-1|x≠1}
