Răspuns:
Explicație pas cu pas:
a) E(x)=[(x³-2x²)/2x²-(x²-4)/4x]:(x-2)/2+1/2=
[x(x²-2x)/2x²-(x²-4)/4x]·2/(x-2)+1/2=
[(x²-2)/2x-(x²-4)/4x]·2/(x-2)+1/2=
{[2(x²-2x)-(x²-4)]/4x}·2/(x-2)+1/2=
[(2x²-4x-x²+4)/4x]·2/(x-2)+1/2=
[(x²-4x+4)/4x]·2/(x-2)+1/2=
[(x-2)²/4x]·2/(x-2)+1/2=2(x-2)²/4x(x-2)+1/2=(x-2)²/2x(x-2)+1/2=
(x-2)/2x+1/2 (x-2)/2x+x/2x=(x-2+x)/2x=(2x-2)/2x=2(x-1)/2x=
(x-1)/x=x/x-1/x=1-1/x
E(x)=1-1/x
b)
E(3)=1-1/3=(3-1)/3=2/3
E(4)=1-1/4=(4-1)/4=3/4
E(5)=1-1/5=)5-1)/5=4/5
E(6)=1-1/6=(6-1)/6=5/6
E(7)=1-1/7=(7-1)/7=6/7
E(8)=1-1/8=(8-1)/8=7/8
E(3)·E(4)·E(5)·E(6)·E(7)·E(8)=
(2/3)·(3/4)·(4/5)·(5/6)·(6/7)·(7/8)=
(2·3·4·5·6·7)/(3·4·5·6·7·8)=2/8=1/4
E(3)·E(4)·E(5)·E(6)·E(7)·E(8)=1/4