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Тамила378
@Тамила378
July 2022
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y^3-(y-4d)(y^2+4yd+16d^2) при y=0,5;d=1
Решите пожалуйста)
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sbornik3HAHUU
Y^3-(y-4d)(y^2+4yd+16d^2) = y^3-y^3-4d^3 = -4d^3 = -4
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Zaznajka2003
Y³-(y-4d)(y²+4yd+16d²) = y³-(y³+4y²d+16yd²-4y²d-16yd²-64d³) = y³-y³-4y²d-16yd²+4y²d+16yd²-64d³ = -64d³
d=1. -64*1=-64
1 votes
Thanks 3
Тамила378
Спасибо)
Zaznajka2003
оххх.... даже не знаю, правильно ли:/
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Answers & Comments
d=1. -64*1=-64