Giải PT $\frac{x+1}{2010}$ + $\frac{x+3}{2008}$+ $\frac{x+4}{2007}$ +$\frac{x+9}{2002}$ = -4

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Giải PT
$\frac{x+1}{2010}$ + $\frac{x+3}{2008}$+ $\frac{x+4}{2007}$ +$\frac{x+9}{2002}$ = -4

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  2. $\frac{x+1}{2010}+$ $\frac{x+3}{2008}+$ $\frac{x+4}{2007}+$ $\frac{x+9}{2002}=-4$

    $=>$$(\frac{x+1}{2010}+1)+$ $(\frac{x+3}{2008}+1)+$ $(\frac{x+4}{2007}+1)+$ $(\frac{x+9}{2002}+1)=-4+4$

    $=>$$\frac{x+2011}{2010}+$ $\frac{x+2011}{2008}+$ $\frac{x+2011}{2007}+$ $\frac{x+2011}{2002}=0$

    $=>(x+2011)$$(\frac{1}{2010}+$ $\frac{1}{2008}+$ $\frac{1}{2007}+$ $\frac{1}{2002})=0$

    $=>x+2011=0$

    $=>x=-2011$

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