[tex]\frac{x^4+ x^3- 5 x^2 -5x}{x^3-x} \\\\x^3-x\neq 0\\\\x(x^2-1)\neq 0\\\\(x-1)(x+1)x\neq 0\\\\x-1\neq 0\ \ i\ \ x+1\neq 0\ \ i\ \ x\neq 0\\\\x\neq 1\ \ i\ \ x\neq -1\\\\D=R\setminus \left\{-1,0,1 \right\}\\\\\frac{x^4+ x^3- 5 x^2 -5x}{x^3-x} =\frac{x^3(x +1)- 5x( x+1)}{x(x^2-1)} =\frac{x^3(x +1)- 5x( x+1)}{x(x^2-1)} =\\\\=\frac{(x^3-5x)(x+1)}{x(x+1)(x-1)}=\frac{(x^3-5x) }{x (x-1)}=\frac{x(x^2-5 ) }{x (x-1)}=\frac{x^2-5}{x-1}[/tex]
[tex]x=\sqrt{3}\\\\\frac{x^2-5}{x-1}=\frac{ (\sqrt{3})^2-5}{ \sqrt{3}-1}=\frac{ 3-5}{ \sqrt{3}-1} =\frac{ -2}{ \sqrt{3}-1}*\frac{\sqrt{3}+1 }{\sqrt{3}+1}=\frac{-2(\sqrt{3}+1}{3-1}=\\\\=\frac{-2(\sqrt{3}+1)}{2}= -(\sqrt{3}+1)[/tex]
