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yachmennikova20
@yachmennikova20
December 2021
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а)1/16-b^2
b) y^2+12y+36
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Shelkovaia
а)1/16-b^2=(1\4-b)(1\4+b) по формуле разности квадратов
b) y^2+12y+36=(у+6)^2=
(у+6)(у+6) по формуле квадрата суммы
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yachmennikova20
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Answers & Comments
b) y^2+12y+36=(у+6)^2=(у+6)(у+6) по формуле квадрата суммы