Question

tìm x biết 

\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)

Answer 1

Bạn sửa đề lại nha.

\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)

=>\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}-3=0\)

=>\(\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1988}-1\right)=0\)

=>\(\frac{x-1-2006}{2006}+\frac{x-10-1997}{1997}+\frac{x-19-1988}{1988}=0\)

=>\(\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)

=>\(\left(x-2007\right).\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)

Vì \(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\ne0\)

=>x-2007=0

=>x=2007