Odpowiedź:
[tex]f(x)=\frac{x^4-x^3-6x^2+6x}{x^3-x}=\frac{x^3(x-1)-6x(x-1)}{x(x^2-1)} =\frac{(x^3-6x)(x-1)}{x(x-1)(x+1)} =\frac{x^3-6x}{x(x+1)}=\frac{x(x^2-6)}{x(x+1)} =\frac{x^2-6}{x+1}[/tex]
gdzie : x≠1
dla [tex]x=\sqrt{7}[/tex] mamy :
[tex]\frac{(\sqrt{7})^2 -6}{\sqrt{7}+1 }=\frac{1}{\sqrt{7}+1 }=\frac{1}{\sqrt{7}+1 } \cdot \frac{\sqrt{7}-1 }{\sqrt{7}-1 } =\frac{\sqrt{7}-1 }{6}[/tex]
