Đáp án:
$x = 2000$
Giải thích các bước giải:
$\frac{x}{2000} + \frac{x+1}{2001} + \frac{x+2}{2002} + \frac{x+3}{2003} = 4 $
$⇔ \frac{x}{2000} + \frac{x+1}{2001} + \frac{x+2}{2002} + \frac{x+3}{2003} - 4 = 0 $
$⇔ ( \frac{x}{2000} - 1 ) + ( \frac{x+1}{2001} - 1 ) + ( \frac{x+2}{2002} - 1 ) + ( \frac{x+3}{2003} - 1 ) = 0 $
$⇔ \frac{x-2000}{2000} + \frac{x-2000}{2001} + \frac{x-2000}{2002} + \frac{x-2000}{2003} = 0 $
$⇔ ( x - 2000 ) . ( \frac{1}{2000} + \frac{1}{2001} + \frac{1}{2002} + \frac{1}{2003} ) = 0 $
mà $\frac{1}{2000} + \frac{1}{2001} + \frac{1}{2002} + \frac{1}{2003} $ khác 0
$⇒ x - 2000 = 0$
$⇔ x = 2000 $
Vậy $ x = 2000$
