Question

Помогите пожалуйста) Срочно...

1) (2y-1) (3y-2)=

2) (1-y)(y+3)=

3) (4a^2+в^2)(3а^2-в^2)=

Answer 1

1) (2y-1)(3y-2)=6y^2-4y-3y+2
2) (1-y)(y+3)=y+3-y^2-3y=-2y+3-y^2=-y^2-2y+3
3) (4a^2+b^2)(3a^2-b^2)=12a^4a^2b^2+3a^2b^2-b^4=12a^4-a^2b^2-b^4

Answer 2

Спс