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zalozenia:
x+1>0 i x+1≠1 i 3x+3>0
D=(-1,0) u (0,+∞)
x²+2x+1-3x-3=0
x²-x-2=0
Δ=1+4*2=9
x=1/2*(1-3)=-1∉D v x=1/2*(1+3)=2
Odp. x=2