rozwiazac nierownosc 1/x(x+1) + 1/(x+1)(x+2) + 1/(x+2)(x+3) +...+1/(x+9)(x+10) < 1/x-2

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;∞)