Comments by "Fhf Fhf" (@fhffhff) on "Бондаренко LIVE" channel.
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1-(x+a+1)²=y,y≤1 2^|x-a|=z≥1[y=z, {y=0,z=1;[{x=-0,5,a=-0,5;{[a=0,a=-1; x=a;{(-0,5;-0,5)(0;0)(-1;-1)}1
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