Đáp án: $x=2000$
Giải thích các bước giải:
Ta có:
$\dfrac{x-1991}{9}+\dfrac{x-1993}{7}+\dfrac{x-1995}{5}+\dfrac{x-1997}{3}+\dfrac{x-1999}{1}=\dfrac{x-9}{1991}+\dfrac{x-7}{1993}+\dfrac{x-5}{1995}+\dfrac{x-3}{1997}+\dfrac{x-1}{1999}$
$\to (\dfrac{x-1991}{9}-1)+(\dfrac{x-1993}{7}-1)+(\dfrac{x-1995}{5}-1)+(\dfrac{x-1997}{3}-1)+(\dfrac{x-1999}{1}-1)=(\dfrac{x-9}{1991}-1)+(\dfrac{x-7}{1993}-1)+(\dfrac{x-5}{1995}-1)+(\dfrac{x-3}{1997}-1)+(\dfrac{x-1}{1999}-1)$
$\to \dfrac{x-1991-9}{9}+\dfrac{x-1993-7}{7}+\dfrac{x-1995-5}{5}+\dfrac{x-1997-3}{3}+\dfrac{x-1999-1}{1}=\dfrac{x-9-1991}{1991}+\dfrac{x-7-1993}{1993}+\dfrac{x-5-1995}{1995}+\dfrac{x-3-1997}{1997}+\dfrac{x-1-1999}{1999}$
$\to \dfrac{x-2000}{9}+\dfrac{x-2000}{7}+\dfrac{x-2000}{5}+\dfrac{x-2000}{3}+\dfrac{x-2000}{1}=\dfrac{x-2000}{1991}+\dfrac{x-2000}{1993}+\dfrac{x-2000}{1995}+\dfrac{x-2000}{1997}+\dfrac{x-2000}{1999}$
$\to (x-2000)(\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{1})=(x-2000)(\dfrac{1}{1991}+\dfrac{1}{1993}+\dfrac{1}{1995}+\dfrac{1}{1997}+\dfrac{1}{1999})$
$\to x-2000=0$
$\to x=2000$
