Thể tích hình hộp chữ nhật:

\(V = \text{d} \overset{ˋ}{\text{a}} \text{i} \times \text{r}ộ\text{ng} \times \text{cao}\)

Vậy:

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

Ta rút gọn:

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

Nên:

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

b,

V=43−4 \(V = 64 - 4\) \(V = 60\)

2x4:x2=2x2

2x2(x2−2)=2x4−4x2

(2x4−3x3−3x2)−(2x4−4x2)=2x4−2x4−3x3−3x2+4x2=
\(- 3 x^{3} + x^{2}\)

−3x3:x2=−3x

−3x(x2−2)=−3x3+6x

(−3x3+x2+6x)−(−3x3+6x)=
\(- 3 x^{3} + 3 x^{3} + x^{2} + 6 x - 6 x\)=
\(x^{2}\)

Hạ tiếp \(- 2\).

x2:x2=1

1(x2−2)=x2−2

(x2−2)−(x2−2)=0=2x2−3x+1

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)

20x3−20x3−10x2+10x2+5x−4x=x

x=−36

a. \(x^{4} - 5 x^{3} + 4 x - 5 - x^{4} + 3 x^{2} + 2 x + 1\)

\(= - 5 x^{3} + 3 x^{2} + 6 x - 4\)

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

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

\(= 2 x^{4} - 5 x^{3} - 3 x^{2} + 2 x - 6\)

hay