Odpowiedzi

2010-04-02T20:35:26+02:00
Warunek x jest różne od -10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,2.
Lewa strona:
1/(x+1)[1/x +1/(x+2)]+1/(x+3)[1/(x+2) +1/(x+4)]+1/(x+5)[1/(x+4)+1/(x+6)]+1/(x+7)[1/(x+6)+1/(x+8)]+1/(x+9)[1/(x+8) +1/(x+10)]= 1/(x+1)[(2x+2)/x(x+2)] + 1/(x+3)[(2x+6)/(x+2)(x+4)]+ 1/(x+5)[(2x+10)/(x+4)(x+6)]+ 1/(x+7)[(2x+14)/(x+6)(x+8)]+ 1/(x+9)[(2x+18)/(x+8)(x+10)] =
2/x(x+2) +2/(x+2)(x+4) + 2/(x+4)(x+6) + 2/(x+6)(x+8) + 2/(x+8)(x+10) =
2/(x+2)[1/x +1/(x+4)] + 2/(x+6)[1/(x+4) +1/(x+8)] + 2/(x+8)(x+10) =
4/x(x+4) + 4/(x+4)(x+8) + 2/(x+8)(x+10)=
4/(x+4) [1/x +1/(x+8)] + 2/(x+8)(x+10)=
8/x(x+8) + 2/(x+8)(x+10)= 1/(x+8) [8/x + 2/(x+10)]=
10/x(x+10).

10/x(x+10)< 1/(x-2)
(10x-20-x²-10x)/[(x-2)x(x+10)]<0
-(x²+10)(x-2)x(x+10)<0
(x²+10)(x-2)x(x+10)>0
x=2 lub x= 0 lub x=-10
Rysujemy wykres tej funkcji i widzimy że wartości większe od 0 przyjmuje dla x∈(-10;0) v (2;∞)