prosze o rozwiazanie nierownosci z wartoscia bezwzgledna: Ix-3/x+1I> badz rowne 7( na dzisiaj);-)

|(x-3)/(x+1)|≥7 (x-3)/(x+1)≥7 |*(x+1)²                      zał. x≠-1 (x-3)(x+1)≥7(x+1)² x²+x-3x-3≥7(x²+2x+1) x²-2x-3≥7x²+14x+7 -6x²-16x-10≥0 -6x²-6x-10x-10≥0 -6x(x+1)-10(x+1)≥0 -(6x+10)(x+1)≥0 6x+10=0 6x=-10 x=-10/6=-5/3 ∨ x=-1 x∈<-5/3,-1>   (x-3)/(x+1)≤-7 |*(x+1)² x²-2x-3≤-7(x²+2x+1) x²-2x-3≤-7x²-14x-7 8x²+12x+4≤0 8x²+8x+4x+4≤0 8x(x+1)+4(x+1)≤0 (8x+4)(x+1)≤0 8x+4=0 8x=-4 x=-1/2 ∨ x=-1 x∈<-1,-1/2>   x∈<-5/3,-1>n<-1,-1/2>{-1} x∈<-5/3,-1)n(-1,-1/2>