Question

[BÙI THỊ NGÂN]

tìm x biết :\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}=\frac{x-4}{2002}\) Giúp mình với 

Answer 1

[ST]

\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}=\frac{x-4}{2002}\)

=>\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}-\frac{x-4}{2004}=0\)

=>\(\left(\frac{x-1}{2005}-1\right)+\left(\frac{x-2}{2004}-1\right)-\left(\frac{x-3}{2003}-1\right)-\left(\frac{x-4}{2002}-1\right)=0\)

=>\(\frac{x-1-2005}{2005}+\frac{x-2-2004}{2004}-\frac{x-3-2003}{2003}-\frac{x-4-2002}{2002}=0\)

=>\(\frac{x-2006}{2005}+\frac{x-2006}{2004}-\frac{x-2006}{2003}-\frac{x-2006}{2002}=0\)

=>\(\left(x-2006\right)\left(\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)

Mà \(\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)

=> x - 2006 = 0 => x = 2006