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標題: 中二恆等式 [打印本頁]

作者: billyconan 時間: 09-9-2 04:31 PM 標題: 中二恆等式

我之前係yahoo到問e條數
Ax(x-1) + B(x+1)(x-1) +C(x+1)x = x^2+4

咁有個人就答
Ax(x-1) + B(x+1)(x-1) +C(x+1)x
=Ax^2 -Ax +Bx^2 -B +Cx^2 +Cx
=(A+B+C)x^2 -(A-C)x -B ≡ x^2 +4

A+B+C = 1 ---(1)
-A +C = 0 ---(2)
-B = 4 ---(3)
in (3), B= - 4
in(1), A - 4 +C = 1
A + C = 5 ---(4)
(4)-(2), 2A=5A = 5/2
in(2), -5/2 +C = 0
C = 5/2
.'. A = 5/2 ,B = - 4 & C = 5/2

我5係好明點解 (4)-(2)會變左2a=5
仲有ar...點減ar???

作者: ChEoNg@@ 時間: 09-9-3 06:09 PM

(4)-(2)
A + C -( -A + C) = 5 - 0
A + C + A - C = 5
2A=5
A= 5/2

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