a)
[tex]\it \dfrac{x}{x^2-x} =\dfrac{\ x^{(x}}{x(x-1)}=\dfrac{1}{x-1}[/tex]
b)
[tex]\it 2+x-2x^2-x^3=x+2-x^2(x+2)=(x+2)(1-x^2)=(x+2)(1-x)(1+x)[/tex]
c)
Folosind a) și b), prima paranteză a expresiei devine:
[tex]\it \dfrac{1}{x-1}+\dfrac{1}{(1-x)(1+x)}+\dfrac{x\cdot x}{x(x+1)}= \dfrac{^{x+1)}1}{x-1}-\dfrac{1}{(x-1)(x+1)}+\dfrac{^{x-1)}x}{x+1}=\\ \\ \\ = \dfrac{x+1-1+x^2-x}{(x-1)(x+1)}=\dfrac{x^2}{x^2-1}\ \ \ \ (1)[/tex]
A doua paranteză a expresiei se scrie:
[tex]\it ^{x)}x-\dfrac{1}{x}= \dfrac{x^2-1}{x}\ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow E(x) =\dfrac{x^2}{x^2-1}\cdot\dfrac{x^2-1}{x} =x[/tex]
