Thể tích hình hộp chữ nhật được tính bằng tích ba kích thước:

\(V = x \left(\right. x + 1 \left.\right) \left(\right. x - 1 \left.\right) .\)

Ta rút gọn:

\(\left(\right. x + 1 \left.\right) \left(\right. x - 1 \left.\right) = x^{2} - 1.\)

Suy ra:

\(V = x \left(\right. x^{2} - 1 \left.\right) = x^{3} - x .\)

Vậy: v=x3-x
V=43−4=64−4=60.

Vậy thể tích khi \(x = 4\) là: 60

Ta có:

\(A = 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2\)
\(B = x^{2} - 2\)

Ta thực hiện phép chia đa thức \(A\) cho \(B\).

2x4:x2=2x2.

\(2 x^{2} \left(\right. x^{2} - 2 \left.\right) = 2 x^{4} - 4 x^{2} .\)

\(\left(\right. 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 \left.\right) - \left(\right. 2 x^{4} - 4 x^{2} \left.\right)\)

\(- 3 x^{3} + x^{2} + 6 x - 2.\)

−3x3:x2=−3x.

\(- 3 x \left(\right. x^{2} - 2 \left.\right) = - 3 x^{3} + 6 x .\)

\(\left(\right. - 3 x^{3} + x^{2} + 6 x - 2 \left.\right) - \left(\right. - 3 x^{3} + 6 x \left.\right)\)

\(x^{2} - 2.\)

x2:x2=1.

\(1 \left(\right. x^{2} - 2 \left.\right) = x^{2} - 2.\)

\(\left(\right. x^{2} - 2 \left.\right) - \left(\right. x^{2} - 2 \left.\right) = 0.\)

Thương là:

Dư là:

Vậy:

5x(4x2−2x+1)−2x(10x2−5x+2)=−36.

5x(4x2−2x+1)=20x3−10x2+5x

\(2 x \left(\right. 10 x^{2} - 5 x + 2 \left.\right) = 20 x^{3} - 10 x^{2} + 4 x .\)

(20x3−10x2+5x)−(20x3−10x2+4x)=−36.

20x3−10x2+5x−20x3+10x2−4x=−36.

(20x3−20x3)+(−10x2+10x2)+(5x−4x)=−36.

\(x = - 36.\)

vậy x=-36

a)Ta có:

\(P \left(\right. x \left.\right) = x^{4} - 5 x^{3} + 4 x - 5\) \(Q \left(\right. x \left.\right) = - x^{4} + 3 x^{2} + 2 x + 1\)

P(x)+Q(x)=(x4−5x3+4x−5)+(−x4+3x2+2x+1)

P(x)+Q(x)=−5x3+3x2+6x−4

b)Ta có:

\(R \left(\right. x \left.\right) = P \left(\right. x \left.\right) - Q \left(\right. x \left.\right)\) \(R \left(\right. x \left.\right) = \left(\right. x^{4} - 5 x^{3} + 4 x - 5 \left.\right) - \left(\right. - x^{4} + 3 x^{2} + 2 x + 1 \left.\right)\)