✔ 最佳答案
1.
Ax(x - 1) + B(x + 1)(x - 1) + C(x + 1)x ≡ x² + 4
Ax² - Ax + B(x² - 1) + C(x² + x) ≡ x² + 4
(A + B + C)x² + (-A + C)x - B ≡ x² + 4
比較常數項:
-B = 4
B = -4
比較 x 項:
-A + C = 0
C = A ...[1]
比較 x² 項:
A + B + C = 1 ... [2]
把 B = -4 及 [1] 代入 [2] 中:
A + (-4) + A = 1
A = 5/2
把 A = 5/2 代入 [1] 中:
C = 5/2
答:A = 5/2, B = -4, C = 5/2
2.
題目中漏了個數字,以「?」表示。
Cx² + A(x + ?) - B ≡ 2(x + 3) + 3(x - 2)
Cx² + Ax + ?A - B ≡ 2x + 6 + 3x - 6
Cx² + Ax + (?A - B) ≡ 5x
比較 x² 項: C = 0
比較 x 項: A = 5
比較常數項: B = ? x A
答: A = 5, B = ? x A, C = 0
3.
a)
左方
= (x - 2)(x + 1)
= x(x + 1) - 2(x + 1)
= x² + x - 2x - 2
= x² - x - 2
= 右方
所以 (x - 2)(x + 1) ≡x² - x - 2
b)
由上: (x - 2)(x + 1) ≡x² - x - 2
令 x = y - 1
[(y - 1) - 2][(y - 1) + 1] ≡(y - 1)² - (y - 1) - 2
(y - 1 - 2)(y - 1 + 1) ≡ (y - 1)² - y + 1 - 2
(y - 3)y ≡ (y - 1)² - y - 1
4.
a)
左方
= (x + 1)(x - 1)
= x(x - 1) + 1(x - 1)
= x² - x + x - 1
= x² - 1
= 右方
b)
由上:(u + 1)(u - 1) ≡ u² - 1
令 u = x²
(x² + 1)(x² - 1) ≡ (x²)² - 1
(x² + 1)(x + 1)(x - 1) ≡ (x²)(x²) - 1