[tex]x^2-4=\frac{2y}{y^2+1}-1=\frac{-(y-1)^2}{2y}=(x-2)(x+2)[/tex]
[tex]x^3+8+3(y-1)^2=0=>-(y-1)^2=\frac{x^3+8}{3}[/tex]
=> [tex]\frac{\frac{x^3+8}{3}}{2y}=(x-2)(x+2)[/tex]
=>[TEX]x+2=0[/TEX]
[tex]\frac{\frac{x^2-2x+4}{3}}{2y}=(x-2)<=>x^2-2x+4=6y(x-2)[/tex]
[tex]=>y=\frac{x^2-2x+4}{6(x-2)}[/tex] (bó tay cái này )