[tex]\it E(x)=\dfrac{(x^2+2x)(x^2+2x+5)+4}{(x^2+2x)(x^2+2x+6)+8}=\dfrac{[(x+1)^2-1][(x+1)^2+4]+4}{[(x+1)^2-1][(x+1)^2+5]+8}\\ \\ \\ Vom\ nota\ (x+1)^2=t, \ iar\ expresia\ devine: \\ \\ \\ E(t)=\dfrac{(t-1)(t+4)+4}{(t-1)(t+5)+8}=\dfrac{t^2+3t-4+4}{t^2+4t-5+8}=\dfrac{t^2+3t}{t^2+4t+3}=\\ \\ \\ =\dfrac{t(t+3)}{(t+1)(t+3)}=\dfrac{t}{t+1}[/tex]
Revenim asupra notației:
[tex]\it E(x)=\dfrac{(x+1)^2}{(x+1)^2+1} \Rightarrow E(x)=\dfrac{x^2+2x+1}{x^2+2x+2}[/tex]
