Nhanh nha,cần gấp lém: Tìm x: $\frac{x+10}{490}$+$\frac{x+20}{480}$+$\frac{x+30}{470}$+$\frac{x+40}{460}$+$\frac{x+50}{4500}$=-5

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1. `(x+10)/(490)+(x+20)/480+(x+30)/470+(x+40)/460+(x+50)/450=-5`

   `(1+(x+10)/490)+(1+(x+20)/480)+(1+(x+30)/470)+(1+(x+40)/460)+(1+(x+50)/450)=0`

   `(x+500)/490+(x+500)/480+(x+500)/470+(x+500)/460+(x+500)/450=0`

   $\left[ \begin{array}{l}x+500=0\\\dfrac{1}{490}+\dfrac{1}{480}+\dfrac{1}{470}+ \dfrac{1}{460}+\dfrac{1}{450}=0\end{array} \right.$

   Ta có `1/490+1/480+1/470+1/460+1/450>0`

   Vậy `x+500=0`

   Suy ra `x=-500`.

   Vậy `x in{-500}`.

    

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2. Đáp án:

    

   Giải thích các bước giải:

    `(x+10)/490 +(x+20)/480 +(x+30)/470 +(x+40)/460 +(x+50)/450=-5`

   `=>(x+10)/490 +(x+20)/480 +(x+30)/470 +(x+40)/460 +(x+50)/450+5=0`

   `=>((x+10)/490 +1)+((x+20)/480 +1)+((x+30)/470 +1)+((x+40)/460 +1) +((x+50)/450+1)=0`

   `=>(x+500)/490 +(x+500)/480 +(x+500)/470 +(x+500)/460+(x+500)/450=0`

   `=>(x+500)(1/490+1/480+1/470+1/460+1/450)=0`

   Ta có:`1/490+1/480+1/470+1/460+1/450>0`

   `=>x+500=0`

   `=>x=-500`

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