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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Giải pt `(x-2)/505 + (x+3)/2015 = (x+6)/1006 + (x+506)/504`

Toán Lớp 8: Giải pt
(x-2)/505 + (x+3)/2015 = (x+6)/1006 + (x+506)/504

Comments ( 2 )

  1. Giải đáp: -2018
    Lời giải và giải thích chi tiết:
    ($\frac{x-2}{505}$+4)+($\frac{x+3}{2015}$+1) =($\frac{x+6}{1006}$+2)+($\frac{x+506}{504}$+3)
    $\frac{x+2018}{505}$+$\frac{x+2018}{2015}$=$\frac{x+2018}{1006}$+$\frac{x+2018}{504}$ 
    $\frac{x+2018}{505}$+$\frac{x+2018}{2015}$-$\frac{x+2018}{1006}$-$\frac{x+2018}{504}$=0
    (x+2018)($\frac{1}{505}$+$\frac{1}{2015}$-$\frac{1}{1006}$-$\frac{1}{504}$=0
    vì $\frac{1}{505}$+$\frac{1}{2015}$-$\frac{1}{1006}$-$\frac{1}{504 > 0
    => x+2018=0 ⇔ x=-2018
     

  2. Giải đáp:
    $x=-2018$
    Lời giải và giải thích chi tiết:
    $\dfrac{x-2}{505}+\dfrac{x+3}{2015}=\dfrac{x+6}{1006}+\dfrac{x+506}{504}$
    $⇔ \dfrac{x-2}{505}+4+\dfrac{x+3}{2015}+1=\dfrac{x+6}{1006}+2+\dfrac{x+506}{504}+3$
    $⇔ \dfrac{x+2018}{505}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{1006}+\dfrac{x+2018}{504}$
    ⇔ (x+2018)(\frac{1}{505}+\frac{1}{2015}-\frac{1}{1006}-\frac{1}{504})=0
    $⇔ \left[ \begin{array}{l}x=-2018(tm)\\\dfrac{1}{505}+\dfrac{1}{2015}-\dfrac{1}{1006}-\dfrac{1}{504}=0(l)\end{array} \right.$
    Vậy $x=-2018$

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222-9+11+12:2*14+14 = ? ( )