Đáp án:
Giải thích các bước giải:
$\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+ \frac{x+4}{2001}$
$⇔ \frac{x+1}{2004}+\frac{x+2}{2003}-\frac{x+3}{2002}- \frac{x+4}{2001}=0$
$⇔\frac{x+1}{2004}+1+\frac{x+2}{2003}+1-\frac{x+3}{2002}-1- \frac{x+4}{2001}-1=0$
$⇔ \frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}- \frac{x+2005}{2001}=0$
$⇔ (x+2005)(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}- \frac{1}{2001})=0$
$⇔ x + 2005 = 0$
$⇔ x = -2005 $
Chúc bn hok tốt @!@@
