Question

\(\frac{y-1}{2004}+\frac{y-2}{2003}-\frac{y-3}{2002}=\frac{y-4}{2001}\)

Answer 1

\(\frac{y-1}{2004}+\frac{y-2}{2003}-\frac{y-3}{2002}=\frac{y-4}{2001}\)

\(\frac{y-1}{2004}-1+\frac{y-2}{2003}-1-\frac{y+3}{2002}+1=\frac{y-4}{2001}-1\)

\(\frac{y-2005}{2004}+\frac{y-2005}{2003}-\frac{y-2005}{2002}=\frac{y-2005}{2001}\)

\(\frac{y-2005}{2001}+\frac{y-2005}{2002}-\frac{y-2005}{2003}-\frac{y-2005}{2004}=0\)

\(\left(y-2005\right).\left(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)

Dễ thấy: \(\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}>0\)

=> y - 2005 = 0

=> y = 2005

Answer 2

Trên thì y - 3 dưới chuyển thành y + 3 v~~