a) Thể tích hình hộp chữ nhật là:

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

Ta rút gọn:

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

nên

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

Vậy:

\(V = x^{3} - x\)

b) Với \(x = 4\):

\(V = 4 \left(\right. 4 + 1 \left.\right) \left(\right. 4 - 1 \left.\right) = 4 \cdot 5 \cdot 3 = 60\)

Kết quả: \(V = 60\)

A(x)=2x4−3x3−3x2+6x−2,B(x)=x2−2

\(\frac{2 x^{4}}{x^{2}} = 2 x^{2}\)

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

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

\(\frac{- 3 x^{3}}{x^{2}} = - 3 x\)

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

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

\(\frac{x^{2}}{x^{2}} = 1\)

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

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

\(\boxed{2 x^{2} - 3 x + 1}\)

5x(4x2−2x+1)−2x(10x2−5x+2)=−36

\(5 x \left(\right. 4 x^{2} - 2 x + 1 \left.\right) = 20 x^{3} - 10 x^{2} + 5 x\)

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

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

Đổi dấu ngoặc:

\(20 x^{3} - 10 x^{2} + 5 x - 20 x^{3} + 10 x^{2} - 4 x\)

Rút gọn:

• \(20 x^{3} - 20 x^{3} = 0\)
• \(- 10 x^{2} + 10 x^{2} = 0\)
• \(5 x - 4 x = x\)

Vậy vế trái còn:

\(x\)

\(x = - 36\)

Kết luận:

\(\boxed{x = - 36}\)

5x(4x2−2x+1)−2x(10x2−5x+2)=−36

\(5 x \left(\right. 4 x^{2} - 2 x + 1 \left.\right) = 20 x^{3} - 10 x^{2} + 5 x\)

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

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

Đổi dấu ngoặc:

\(20 x^{3} - 10 x^{2} + 5 x - 20 x^{3} + 10 x^{2} - 4 x\)

Rút gọn:

• \(20 x^{3} - 20 x^{3} = 0\)
• \(- 10 x^{2} + 10 x^{2} = 0\)
• \(5 x - 4 x = x\)

Vậy vế trái còn:

\(x\)

\(x = - 36\)

Kết luận:

\(\boxed{x = - 36}\)