[tex]E(x)=(\frac{x+2}{x-5}-\frac{x-5}{x+2})\cdot \frac{x^2-4x-5}{2x^2-x-3}\\
Descompunem\ numitorii\ separat:\\
x^2-4x-5=x^2-5x+x-5=x(x-5)+(x-5)=(x+1)(x-5)\\
2x^2-x-3=2x^2+2x-3x-3=2x(x+1)-3(x+1)=\\
=(x+1)(2x-3)\\
Revenind:\\
E(x)=\frac{(x+2)^2-(x-5)^2}{(x-5)(x+2)}\cdot \frac{(x+1)(x+5)}{(x+1)(2x-3)}\\
E(x)=\frac{(x+2-x+5)(x+2+x-5)}{(x+2)(2x-3)}\\
E(x)=\frac{7(2x-3)}{(x+2)(2x-3)}\\
\boxed{\bold{E(x)=\frac{7}{2x-3},\forall\ x\in R-\{-2,-1,\frac{3}{2},5\}}}}[/tex]
