a) thể tích=tích 3 kích thước

V=x(x+1)(x-1)

Rút gọn:

V=x×(x^2-1)=x^3-x

b) Thay x=4

V=x(x+1)(x-1)

=4×(4+1)×(4-1)

=4×5×3

=60

5x(4x^2-2x+1)-2x(10x^2-5x+2)=-36

X=(20x^3-10x^2+5x)-(20x^3-10x^2+4x)=-36

=20x^3-10x^2+5x-20x^3+10x^2-4x=-36

=(20x^3-20x^3)+(-10x^2+10x^2)+(5x-4x)=-36

X=-36

P(x)+Q(x)=(x^4-5x^3+4x-5)+(-x^4+3x^2+2x+1)

=x^4-5x^3+4x-5-x^4+3x^2+2x+1

=(x^4-x^4)-5x^3+3x^2+(4x+2x)+(-5+1)

=0x^4-5x^3+6x+3x^2-4

Vậy P(x)+Q(x)=-5x^3+3x^2+6x-4

b)Từ P(x)=R(x)+Q(x),suy ra R(x)=P(x)-Q(x)

R(x)=(x^4-5x^3+4x-5)-(-x^4+3x^2+2x+1)

=x^4-5x^3+4x-5+x^4-3x^2-2x-1

=(x^4+x^4)-5x^3-3x^2+(4x-2x)+(-5-1)

=2x^4-5x^3-3x^2+2x-6

Vậy R(x)=2x^4-5x^3-3x^2+2x-6