解方程式：3/x + 1/(x-1) +...

Question

解方程式：3/x + 1/(x-1) + 4/(x-2) + 4/(x-3) + 1/(x-4) + 3/(x-5) = 0

答案有五個解哦！

Answer 1

x=5/2 or (5±√7)/2 or (5±√17)/2
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3/x + 1/(x-1) + 4/(x-2) + 4/(x-3) + 1/(x-4) + 3/(x-5) = 0
3/x + 3/(x-5) + 1/(x-1) + 1/(x-4) + 4/(x-2) + 4/(x-3) = 0
3(2x-5)/[x(x-5)] + (2x-5)/[(x-1)(x-4)] + 4(2x-5)/[(x-2)(x-3)] = 0
(2x-5)[ 3/(x^2-5x) + 1/(x^2-5x+4) + 4/(x^2-5x+6) ] = 0
故其中一解是 2x-5=0 ，即 x=5/2
令 x^2-5x = y
3/(x^2-5x) + 1/(x^2-5x+4) + 4/(x^2-5x+6) = 0
3/y + 1/(y+4) + 4/(y+6) = 0
3(y+4)(y+6) + y(y+6) + 4y(y+4) = 0
8y^2 + 52y + 72 = 0
2y^2 + 13y + 18 = 0
(2y+9)(y+2) = 0
y = -9/2 or -2
x^2 - 5x + 25/4 = y + 25/4
(x - 5/2)^2 = 7/4 or 17/4
x= (5±√7)/2 or (5±√17)/2

Answer 2

首先,可以發現分子,他的分子有兩個3,兩個1,兩個4..把他們兩個兩個並在一起,就是....3/x和3/(x-5);1/(x-1)和1/(x-4);4/(x-2)和4/(x-3)....然後分別通分,嘿嘿～你就會發現...他們的分母都是(2x-5)...因為要=0(分子可以當零,分母不行),所以答案就是5/2

2005-11-17 20:01:00 補充：
啥?對不起...我忘了...把我的當笑話看吧..我以為很簡單就寫的很爽,另一位大大寫的很好..雖然有點麻煩,但應該就是這樣沒錯..