Question

tìm x : x-1/2004+x-2/2003=x-3/2002+x-4/2001

Answer 1

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

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

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

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

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

Vì \(\frac{1}{2001}>\frac{1}{2003};\frac{1}{2002}>\frac{1}{2004}\)

\(\Rightarrow\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\ne0\)

\(\Rightarrow x-2005=0\)

\(\Rightarrow x=2005\)

Answer 2

x+2005=0