[latex]\bf \cfrac{2}{x}+\cfrac{x}{x+1}+\cfrac{x^2}{x-1}\impliedby \begin{array}{llll} \textit{now, our LCD is}\\ x(x+1)(x-1) \end{array} \\\\\\ \cfrac{[2~(x+1)(x-1)]~+~[x~[x(x-1)]]~+~[x^2~[x(x+1)]]}{x(x+1)(x-1)} \\\\\\ \cfrac{[2x^2-2]~+~[x^3-x^2]~+~[x^4+x^3]}{x(x+1)(x-1)} \\\\\\ \cfrac{2x^2-2~+~x^3-x^2~+~x^4+x^3}{x(x+1)(x-1)} \implies \cfrac{x^4+2x^3-x^2-2}{x(x+1)(x-1)}[/latex] [latex]\bf \textit{and now, if you recall }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -------------------------------\\\\ \textit{then we can combine }(x+1)(x-1)\implies x^2-1 \\\\\\ \cfrac{x^4+2x^3-x^2-2}{x(x+1)(x-1)}\implies \cfrac{x^4+2x^3-x^2-2}{x(x^2-1)}\implies \cfrac{x^4+2x^3-x^2-2}{x^3-x}[/latex]

hello here is a solution : the commun denominateur is : x(x-1)(x+1) ....continue