Question

$E(x)=( \frac{x+1}{x}- \frac{x}{x+1}): ( \frac{1}{ x^{2} } - \frac{1}{(x+1)^{2} })$,unde x apartine R \{-1,0}
Se considera expresia [tex]E(x)=( \frac{2x}{3x}- \frac{6}{x}): ( \frac{x^{2}-4x+3 }{ x^{2} -x} - 2)
[/tex],unde x apartine R\ {0,1,3}.Aratati ca [tex]E(x)= \frac{2x}{3}
[/tex] oricare ar fi x apartine R\ {0,1,3}

Answer 1

E(x)={[(x+1)(x+1)-x²]/[x(x+1)]}:{[(x+1)²-x²]/[x²(x+1)²]}
E(x)=[(x²+x+x+1-x²)/x(x+1)]:[(x²+2x+1-x²)/x²(x+1)²]
E(x)=[(2x+1)/x(x+1)]:[(2x+1)/x²(x+1)²]
E(x)=[(2x+1)/x(x+1)]·[x²(x+1)²/(2x+1)] simplificam cu 2x+1 si cu  si x+1 rezulta
E(x)=x(x+1)=x²+x