|x+2|/(2x²-x-3)>|x+2|/(x²+x-20)
|x+2|/(2x²-x-3)-|x+2|/(x²+x-20)>0
|x+2|*(1/(2*(x+1)(x-1,5))-1/((x+5)(x-4)))>0
ОДЗ: x+2≠0 x≠-2 x+1≠0 x≠-1 x-1,5≠0 x≠1,5 x+5≠0 x≠-5 x-4≠0 x≠4
|x+2|*(x²+x-20-2x²+x+3)/(2*(x+1)(x-1,5)(x+5)(x-4))>0
|x+2|*(-x²+2x-17)/(2*(x+1)(x-1,5)(x+5)(x-4))>0
|x+2|*(-(x²-2x+17))/(2*(x+1)(x-1,5)(x+5)(x-4))>0 |÷(-1)
|x+2|*(x²-2x+1+16)/(2*(x+1)(x-1,5)(x+5)(x-4))<0<br>|x+2|*((x-1)²+16)/(2*(x+1)(x-1,5)(x+5)(x-4))<0 |×2<br>|x+2|*((x-1)²+16)/((x+1)(x-1,5)(x+5)(x-4))<0<br>Так как x+2>0, если х≠-2, а (x-1)²+16>0 ⇒
-∞___+___-5___-___-2___-___-1___+___1,5____-____4____+____+∞
x∈(-5;-2)U(-2;-1)U(1,5;4) ⇒
∑=-4+(-3)+2+3=-7+5=-2.
Ответ: ∑=-2.