Re: [中學] 分式方程式
※ 引述《jinnawang (新的開始)》之銘言：
: (x-1/x-2)-(x-2/x-3)-(x-3/x-4)+(x-4/x-5)=0
: 請問大大們
: 除了通分
: 應該怎麼解呢?
(x-1/x-2)-(x-2/x-3)-(x-3/x-4)+(x-4/x-5)
= (1+ 1/x-2)-(1+ 1/x-3)-(1+ 1/x-4)+(1+ 1/x-5)
= (1/x-2 + 1/x-5) - (1/x-3 + 1/x-4)
= [(x-7)/(x-2)(x-5)] - [(x-7)/(x-3)(x-4)]
= (x-7)[1/(x-2)(x-5) - 1/(x-3)(x-4)]
= (x-7)[2/(x-2)(x-5)(x-3)(x-4)]
= 2(x-7)/(x-2)(x-5)(x-3)(x-4)
= 0
=> x=7
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