Question

Se consideră expresia E(x)=(x+1/x-x/x+1):(1/x^2-1/(x+1)^2),unde x apartine R\{-1;0}.Demonstrati ca,pt.orice nr nat nenul n,E(n) este nr natural par.

Answer 1

[tex]E(x)=(\frac{x+1}{x}-\frac{x}{x+1}):(\frac{1}{x^2}-\frac{1}{(x+1)^2})\\ E(x)=\frac{(x+1)^2-x^2}{x(x+1)}:\frac{(x+1)^2-x^2}{x^2\cdot (x+1)^2}\\ E(x)=\frac{(x+1)^2-x^2}{x(x+1)}\cdot \frac{x^2\cdot (x+1)^2}{(x+1)^2-x^2}\\ E(x)=x(x+1)\forall x\in R - \{0,-1\}\\ Deci\ E(x)\ este\ numar\ par.[/tex]