(3d-2)(3-2d)+(2d-3) при d = 3,2

Question

(3d-2)(3-2d)+(2d-3) при d = 3,2

Answer 1

(3d-2)(3-2d)+(2d-3)=(3d-2)(3-2d)-(3-2d).1=
=(3-2d)(3d-2-1)=(3-2d)(3d-3)=3(d-1)(3-2d)
d=3,2
3.(3,2-1)(3-6,4)=3.2,2.(-3,4)=-6,6.3,4=-22,44

Answer 2

(3-2d)(3d-2-1)=(3-2d)(3d-3)=3(d-1)(3-2d)=3(3.2-1)(3-2*3.2)=3*2.1*(3-6.4)=
6.3*(-3.4)=-21.42