`a)`\(B=\left(\dfrac{x-2}{x+1}+\dfrac{x+2}{1-x}+\dfrac{6x^2}{x^2-1}\right).\dfrac{x^3-x}{x-1}\);\(x\ne\pm1\)
\(B=\left(\dfrac{\left(x-2\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+6x^2}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{x\left(x^2-1\right)}{x-1}\)
\(B=\dfrac{x^2-x-2x+2-x^2-x-2x-2+6x^2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x\left(x-1\right)\left(x+1\right)}{x-1}\)
\(B=\dfrac{6x^2-6x}{\left(x-1\right)\left(x+1\right)}.x\left(x+1\right)\)
\(B=\dfrac{6x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}.x\left(x+1\right)\)
\(B=6x^2\)
`b)`\(B=2x^2\)
\(\Leftrightarrow6x^2=2x^2\)
\(\Leftrightarrow x=0\left(tm\right)\)
Vậy \(S=\left\{0\right\}\)
