Question

se consideră expresia e (x) =​

Answer 1

$\displaystyle E(x)=\frac{x}{x^2+x}-\left(\frac{x}{x-1}-\frac{x}{x+1}\right):\frac{2x}{x-1}=\\\\=\frac{x}{x(x+1)}-\frac{x(x+1)-x(x-1)}{(x-1)(x+1)}\cdot\frac{x-1}{2x}=\\\\=\frac{1}{x+1}-\frac{x^2+x-x^2+x}{x+1}\cdot\frac1{2x}=\\\\=\frac{1}{x+1}-\frac{2x}{x+1}\cdot\frac1{2x}=\frac{1}{x+1}-\frac1{x+1}=0$