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作者: zenka 時間: 09-4-15 04:21 PM 標題: 方程~~急

(2bx+2cx-2bc-2b^2)/(bc+cx+-ab-ac)-(ax-cx-a^2+ac)/(ax-cx-ab+bc)= 1

x是未知數,找x。

作者: Naozumi 時間: 09-4-15 11:08 PM

(2bx+2cx-2bc-2b^2)/(bc+cx+-ab-ac) = [2x(b+c) -2b(c+b)]/[x(b+c) - a(b+c)]
                                                      = 2(b+c)(x-b)/[(b+c)(x-a)]
                                                      = 2(x-b)/(x-a)

(ax-cx-a^2+ac)/(ax-cx-ab+bc) = [x(a-c)-a(a-c)]/[x(a-c)+b(c-a)]
                                             = (a-c)(x-a)/[(a-c)(x-b)]
                                             = (x-a)/(x-b)

2(x-b)/(x-a) - (x-a)/(x-b) =1
[2(x-b)^2 - (x-a)^2]/[(x-a)(x-b)]=1
2(x-b)^2 - (x-a)^2 = (x-a)(x-b)
2x^2 -4bx +2b^2 -x^2 + 2ax -a^2 = x^2 -(a+b)x + ab
x^2 -(4b - 2a)x + 2b^2 - a^2 = x^2 -(a+b)x + ab
(3b - 3a)x = 2b^2 - a^2 - ab
3(b - a)x= (b^2 - a^2) + (b^2 - ab)
3(b - a)x= (b -a )(b+a) + b (b - a)
x = (a + 2b)/3

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