(x-2x2-1-x+2x2+2x+1)⋅(1-x22)2=0(x≠±1)
⇔[x-2(x-1)(x+1)-x+2(x+1)2]⋅(x2-1)24=0
⇔[(x-2)(x+1)(x-1)(x+1)2-(x+2)(x-1)(x-1)(x+1)2]⋅(x2-1)24=0
⇔x2-x-2+x2-x+2(x-1)(x+1)2⋅(x-1)2⋅(x+1)24=0
⇔2x2-2x(x-1)(x+1)2⋅(x-1)2⋅(x+1)24=0
⇔(2x2-2x)(x-1)2=0
⇔(2x2-2x)(x-1)=0
⇔2x(x-1)2=0
⇔x=0(n) hoặc x=1(l)
→S={0}
#2