數學暑期作業中有題唔識幫幫手

2010-07-30 8:31 am
若(X+1)2+A(2X+3)≡X2+2BX+4B,求A及B的值
提示:建立以A及B為變數的聯立方程

回答 (1)

2010-07-30 9:39 am
✔ 最佳答案
1.
(x + 1)² + A(2x + 3) ≡ x² + 2Bx + 4B
x² + 2x + 1 + 2Ax + 3A ≡ x² + 2Bx + 4B
x² + (2A + 2)x + (3A + 1) ≡ x² + 2Bx + 4B

比較 x 項: 2A + 2 = 2B …… (1)
比較常數項: 3A + 1 = 4B …… (2)

(1) x 2:
4A + 4 = 4B …… (3)

(3) = (2)
4A + 4 = 3A + 1
A = -3

把A = -3 代入 (1) 中:
2(-3) + 2 = 2B
2B = -4
B = -2

所以 A = -3,B = -2


2.
(a - b)(a + b)(a² + b²)(a⁴ + b⁴)(a⁸ + b⁸)(a¹⁶ + b¹⁶)(a³² + b³²)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a² - b²)(a² + b²)(a⁴ + b⁴)(a⁸ + b⁸)(a¹⁶ + b¹⁶)(a³² + b³²)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a⁴ - b⁴)(a⁴ + b⁴)(a⁸ + b⁸)(a¹⁶ + b¹⁶)(a³² + b³²)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a⁸ - b⁸)(a⁸ + b⁸)(a¹⁶ + b¹⁶)(a³² + b³²)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a¹⁶ - b¹⁶)(a¹⁶ + b¹⁶)(a³² + b³²)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a³² - b³²)(a³² + b³²)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a⁶⁴ - b⁶⁴)(a⁶⁴ + b⁶⁴)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a¹²⁸ - b¹²⁸)(a¹²⁸ + b¹²⁸)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a²⁵⁶ - b²⁵⁶)(a²⁵⁶ + b²⁵⁶)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a⁵¹² - b⁵¹²)(a⁵¹² + b⁵¹²)(a¹⁰²⁴ + b¹⁰²⁴)
= (a¹⁰²⁴ - b¹⁰²⁴)(a¹⁰²⁴ + b¹⁰²⁴)
= a²⁰⁴⁸ - b²⁰⁴⁸
參考: andrew


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