Đáp án:
$x=-11$
Giải thích các bước giải:
$\frac{x+5}{2006}+\frac{x+4}{2007}+\frac{x+3}{2008}=\frac{x+9}{2002}+\frac{x+10}{2001}+\frac{x+11}{2000}\Leftrightarrow \frac{x+5}{2006}+1+\frac{x+4}{2007}+1+\frac{x+3}{2008}+1=\frac{x+9}{2002}+1+\frac{x+10}{2001}+1+\frac{x+11}{2000}+1\Leftrightarrow \frac{x+2011}{2006}+\frac{x+2011}{2007}+\frac{x+2011}{2008}=\frac{x+2011}{2002}+\frac{x+2011}{2001}+\frac{x+2011}{2000}\Rightarrow (x+11)(\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2002}-\frac{1}{2001}-\frac{1}{2000})=0\Rightarrow x+11=0\Rightarrow x=-11 $ (Vì $(...)$ luôn khác $0$)
Vậy $x=-11$
