Question

(d−5)(7d+1)(3d−7).







Приклад:
a)(7+b)⋅(4a5+2ab)=7⋅4a5+7⋅2ab+b⋅4a5+b⋅2ab==28a5+14ab+4a5b+2ab2.


б)(5xy−3x2)⋅(x2+1)=5xy⋅x2+5xy⋅1−3x2⋅x2−3x2⋅1==5x1+2y+5xy−3x2+2−3x2=5x3y+5xy−3x4−3x2.

Answer 1

Ответ:

$21d^3 - 151d^2 + 223d + 35$

Пошаговое объяснение:

$(d-5)(7d+1)(3d-7)=(d\cdot 7d+d\cdot 1-5\cdot 7d-5\cdot 1)(3d-7)=\\\\(7d^2+d-35d-5)(3d-7)=(7d^2-34d-5)(3d-7)=\\\\7d^2\cdot 3d-7d^2\cdot 7-34d\cdot 3d+34d\cdot7-5\cdot 3d+5\cdot 7=\\\\21d^3-49d^2-102d^2+238d-15d+35=\\\\21d^3 - 151d^2 + 223d + 35$