Question

exercitiul 5 va rog. :(

Answer 1

$E(x)=(1+ \frac{2-x}{x+1}) : \frac{x-1}{(2x+1) ^{2}-(x+2) ^{2} }$
$E(x)=(\frac{x+1+2-x}{x+1}) * \frac{(2x+1) ^{2}-(x+2) ^{2} }{x-1}$
$E(x)=(\frac{3}{x+1}) * \frac{(2x+1-x-2) (2x+1+x+2)}{x-1}$
$E(x)=(\frac{3}{x+1}) * \frac{(x-1) (3x+3)}{x-1}$
$E(x)=\frac{3}{(x+1)} * \frac{3(x-1) (x+1)}{x-1}$
$E(x)=9$