x-11 / 111 + x-12 / 112 = x -23 / 123 + x -24 /124 tìm x

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1. `(x-11)/111+(x-12)/112=(x-23)/123+(x-24)/124`

   `⇔(x-11)/111+1+(x-12)/112+1=(x-23)/123+1+(x-24)/124+1`

   `⇔(x-11+111)/111+(x-12+112)/112=(x-23+123)/123+(x-24+124)/124`

   `⇔(x+100)/111+(x+100)/112=(x+100)/123+(x+100)/124`

   `⇔(x+100).(1/111+1/112-1/123-1/124)=0`

   `⇔x+100=0` `(` vì `1/111+1/112-1/123-1/124\ne0)`

   `⇔x=-100`

   Vậy `S={-100}`

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2. $\dfrac{x-11}{111}+\dfrac{x-12}{112}=\dfrac{x-23}{123}+\dfrac{x-24}{124}$

   $\to \dfrac{x-11}{111}+1+\dfrac{x-12}{112}+1=\dfrac{x-23}{123}+1+\dfrac{x-24}{124}+1$

   $\to \dfrac{x+100}{111}+\dfrac{x+100}{112}=\dfrac{x+100}{123}+\dfrac{x+100}{124}$

   $\to \dfrac{x+100}{111}+\dfrac{x+100}{112}-\dfrac{x+100}{123}-\dfrac{x+100}{124}=0$

   $\to (x+100) \left(\dfrac{1}{111}+\dfrac{1}{112}-\dfrac{1}{123}-\dfrac{1}{124} \right)=0$

   Vì $\dfrac{1}{111}>\dfrac{1}{123}\to \dfrac{1}{111}-\dfrac{1}{123}>0$

   $\dfrac{1}{112}>\dfrac{1}{124}\to \dfrac{1}{112}-\dfrac{1}{124}>0$

   $\to \dfrac{1}{111}+\dfrac{1}{112}-\dfrac{1}{123}-\dfrac{1}{124}>0$

   $\to x+100=0$

   $\to x=-100$

   Vậy $x=-100$

    

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