Question

Calculati E(n) va rog frumos dau puncte multe si cel mai bun raspuns!!!!![tex]E{(n)}= \frac{x^{2}+4x+4}{x(x+2)} : (1+ \frac{2}{x});
E{(n)}= (\frac{x-2}{x^{2}-4} * \frac{5x+10}{x-3} + 1) * \frac{x-3}{x+2}
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Answer 1

E(n)={(x^2+4x+4)/[x(x+2)]}:(1+ 2/x)=

={[(x+2)^2]/[x(x+2)]}:[(x+2)/x]=

=[(x+2)/x]*[x/(x+2)]=[x(x+2)]/[x(x+2)]=1


E(n)={[(x-2)/x^2 -4)]*[(5x+10)/(x-3)] +1}*[(x-3)/(x+2)]=

={(x-2)/[(x-2)(x+2)] *5[(x+2)/(x-3)] +1}*[(x-3)/(x+2)]=

=[1/(x+2)]*5[(x+2)/(x-3)] +1}*[(x-3)/(x+2)]=

=[5/(x-3) +1]*[(x-3)/(x+2)]=

=[(5+x-3)/(x-3)]*[(x-3)/(x+2)]=

=[(x+2)/(x-3)]*[(x-3)/(x+2)]=

[(x+2)(x-3)]/[(x-3)(x+2)]=1