Răspuns:
Explicație pas cu pas:
Fie x²+x=t, atunci (x² + x)(x² + x-1)-2=t(t-1)-2=t²-t-2=t²+2t+1-2t-t-1-2=(t+1)²-3(t+1)=(t+1)(t+1-3)=(t+1)(t-2)
(x² + x)(x² + x + 2)+1=t(t+2)+1=t²+2t+1=(t+1)²
Deci (t+1)(t-2)/(t+1)²=(t-2)/(t+1).
Revenim la variabila x, ⇒(x²+x-2)/(x²+x+1)
[tex]\it F(x)=\dfrac{(x^2+x)(x^2+x-1)-2}{(x^2+x)(x^2+x+2)+1}\\ \\ \\ Vom\ nota\ x^2+x=y,\ iar\ frac\c{\it t}ia\ devine:\\ \\ \\ F(y)=\dfrac{y(y-1)-2}{y(y+2)+1}=\dfrac{y^2-y-2}{y^2+2y+1}=\dfrac{y^2-2y+y-2}{(y+1)^2}=\dfrac{y(y-2)+(y-2)}{(y+1)^2}=\\ \\ \\ =\dfrac{(y-2)(y+1)}{(y+1)(y+1)}=\dfrac{y-2}{y+1}[/tex]
Revenind asupra notației, obținem:
[tex]\it F(y)=\dfrac{x^2+x-2}{x^2+x+1}[/tex]
