Giai pt 1)x+1/2030+x+2/2019=x+3/3018+x+4/2017 2)59-x/41+57-x/43+55-x/45+53-x/47=-4 3)x-3/x+3-x+3/x-3=4x^2/9-x^2

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1. Đáp án:

    

   Giải thích các bước giải:

    1 )   ( x+1 / 2020    +1 ) + ( x+2 /2019     +1 )  = (  x+3 /2018    +1 ) + ( x+4 /2017     +1 )

         <=>  x+2021 /2020  +  x+2021 / 2019 –  x+2021 /2018 – x+2021 /2017 =0

         <=> (x+2021) +1/2020 +1/2019 – 1/2018 – 1/2017 =0

          <=> x+2021=0 

          <=> x= -2021

   2) <=> x+100/ 41 + x+100/43 + x+100/45 + x+100/47   =0

       <=> (x+100)  + 1/41 + 1/43 + 1/45+ 1/47  =0 

       <=> x+100 =0 

       <=> x = -100

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2. `1)(x+1)/2020+(x+2)/2019=(x+3)/2018+(x+4)/2017`

   `⇔(x+1)/2020+1+(x+2)/2019+1=(x+3)/2018+1+(x+4)/2017+1`

   `⇔(x+1+2020)/2020+(x+2+2019)/2019=(x+3+2018)/2018+(x+4+2017)/2017`

   `⇔(x+2021)/2020+(x+2021)/2019=(x+2021)/2018+(x+2021)/2017`

   `⇔(x+2021)(1/2020+1/2019-1/2018-1/20017)=0`

   `⇔x+2021=0`

    `⇔x=-2021`

   `⇒` Vậy `S={-2021}`

   2)`(59-x)/41+(57-x)/43+(55-x)/45+(53-x)/47=0`

   `⇔(59-x)/41+1+(57-x)/43+1+(55-x)/45+1++(53-x)/47+1=0`

   `⇔(59-x+41)/41+(57-x+43)/43+(55-x+45)/45+(53-x+47)/47=0`

   `⇔(100-x)/41+(100-x)/43+(100-x)/45+(100-x)/47=0`

   `⇔(100-x)(1/41+1/43+1/45+1/47)=0`

   `⇔100-x=0`

   `⇔x=-100`

   `⇒`Vậy `S={-100}`

   `3)(x-3)/(x+3)-(x+3)/(x-3)=(4x^2)/(9-x^2)`

   `⇔(x-3)/(x+3)-(x+3)/(x-3)=(-4x^2)/(x^2-3^2)`

   `ĐKXĐ:(x+3)(x-3)\ne0`

   $x\neq±3$

   `⇔(x-3)^2/((x+3)(x-3)) – (x+3)^2/((x+3)(x-3))=(-4x^2)/((x+3)(x-3))`

   `⇔x^2-6x+9-(x^2+6x+9)=-4x^2`

   `⇔x^2-6x+9-x^2-6x-9+4x^2=0`

   `⇔-12x+4x^2=0`

   `⇔4x(-3+x)=0`

   \(\left[ \begin{array}{l}4x=0\\-3+x=0\end{array} \right.\) 

   `⇒x=0(nhận)`

   `⇒x=3(loại)`

   `⇒` Vậy `S={3}`

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