Ответ №1
(2b²-b)/[(b+1)(b²-b+1)]-(b-1)/(b²-b+1)=
=[(2b²-b-(b+1)(b-1)]/[(b+1)(b²-b+1)]=
(2b²-b-b²+1)/[(b+1)(b²-b+1)]=(b²-b+1)/[(b+1)(b²-b+1)]=1/(b+1)
Ответ №1
(2b²-b)/[(b+1)(b²-b+1)]-(b-1)/(b²-b+1)=
=[(2b²-b-(b+1)(b-1)]/[(b+1)(b²-b+1)]=
(2b²-b-b²+1)/[(b+1)(b²-b+1)]=(b²-b+1)/[(b+1)(b²-b+1)]=1/(b+1)
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