Question

Pls help. Chiar nu stiu cum se face. Stau de 10 minutea ea :((. Mi-a dat gresit. ​

Answer 1

$\displaystyle\it\\E(x)=\Bigg(\frac{1}{x^2-x}-\frac{1}{x^2+x}\Bigg)\cdot\frac{x^3-x}{x+2}.\\E(x)=\Bigg(\frac{1}{x(x-1)}-\frac{1}{x(x+1)}\Bigg)\cdot\frac{x(x^2-1)}{x+2}.\\E(x)=\Bigg(\frac {x+1}{(x-1)x(x+1)}-\frac{x-1}{(x-1)x(x+1)}\Bigg)\cdot \frac{x(x^2-1)}{x+2}.\\E(x)=\Bigg( \frac{2}{(x-1)x(x+1)}\Bigg)\cdot\frac{(x-1)x(x+1)}{x+2}.\\\boxed{\it E(x)=\frac{2}{x+2},~\forall~x\in\left\{-2,-1,0,1\right\}}.$

Answer 2

Răspuns:

E(x)=2/x+2

Explicație pas cu pas:

E(x)=[1/x•(x-1)-1/x•(x+1)]•x•(x-1)•(x+1)/x+2,unde x=R\{-2,-1,0,1};

E(x)=(x+1-x+1/x^3-x)•(x^3-x)/x+2

=>E(x)=2/x+2