Thể tích hình hộp chữ nhật

⇒ V = x(x + 1)(x − 1)

Ta có:

(x + 1)(x − 1) = x² − 1

Nên:

V = x(x² − 1)
V = x³ − x

2x² − 3x + 1
x² − 2 ) 2x⁴ − 3x³ − 3x² + 6x − 2
2x⁴ − 4x²
-------------------
−3x³ + x² + 6x
−3x³ + 6x
----------------
x² − 2
x² − 2
----------------
0

5x(4x2−2x+1)=20x3−10x2+5x \(2 x \left(\right. 10 x^{2} - 5 x + 2 \left.\right) = 20 x^{3} - 10 x^{2} + 4 x\)

\(20 x^{3} - 10 x^{2} + 5 x - \left(\right. 20 x^{3} - 10 x^{2} + 4 x \left.\right) = - 36\)

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

\(x = - 36\)

Vậy nghiệm của phương trình là:

\(x = - 36\)

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

Gộp các hạng tử cùng bậc:

\(= x^{4} - x^{4} - 5 x^{3} + 3 x^{2} + 4 x + 2 x - 5 + 1\) \(= - 5 x^{3} + 3 x^{2} + 6 x - 4\)

Vậy:\(P \left(\right. x \left.\right) + Q \left(\right. x \left.\right) = - 5 x^{3} + 3 x^{2} + 6 x - 4\)

Ta có:

\(R \left(\right. x \left.\right) = P \left(\right. x \left.\right) - Q \left(\right. x \left.\right)\)\(R \left(\right. x \left.\right) = \left(\right. x^{4} - 5 x^{3} + 4 x - 5 \left.\right) - \left(\right. - x^{4} + 3 x^{2} + 2 x + 1 \left.\right)\)\(= x^{4} - 5 x^{3} + 4 x - 5 + x^{4} - 3 x^{2} - 2 x - 1\) \(= 2 x^{4} - 5 x^{3} - 3 x^{2} + 2 x - 6\)

Vậy:

\(R \left(\right. x \left.\right) = 2 x^{4} - 5 x^{3} - 3 x^{2} + 2 x - 6\)