Salut,
[tex]E(x)=\dfrac{x}{x^2+x}-\left(\dfrac{x}{x-1}-\dfrac{x}{x+1}\right):\dfrac{2x}{x-1}=\\\\\\=\dfrac{x}{x(x+1)}-\dfrac{x(x+1)-x(x-1)}{(x-1)(x+1)}\cdot\dfrac{x-1}{2x}=\\\\\\=\dfrac{1}{x+1}-\dfrac{x^2+x-x^2+x}{x+1}\cdot\dfrac{1}{2x}=\\\\\\=\dfrac{1}{x+1}-\dfrac{2x}{x+1}\cdot\dfrac{1}{2x}=\dfrac{1}{x+1}-\dfrac{1}{x+1}=0.[/tex]
Ai înțeles rezolvarea ?
Green eyes.