Giải thích các bước giải:
Ta có :
$\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}$
$\to (\dfrac{x-10}{1994}-1)+(\dfrac{x-8}{1996}-1)+(\dfrac{x-6}{1998}-1)+(\dfrac{x-4}{2000}-1)=(\dfrac{x-2002}{2}-1)+(\dfrac{x-2000}{4}-1)+(\dfrac{x-1998}{6}-1)+(\dfrac{x-1996}{8}-1)+(\dfrac{x-1994}{10}-1)$
$\to\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}$
$\to (x-2004)(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000})=(x-2004)(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10})$
$\to (x-2004)(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10})=0$
$\to x-2004=0$
$\to x=2004$
