Question

Giải phương trình:

\(\frac{x-1}{2009}\) + \(\frac{x-2}{2010}\) = \(\frac{x+5}{2003}\) + \(\frac{x+7}{2001}\)

Answer 1

\(\frac{x-1}{2009}+\frac{x-2}{2010}=\frac{x+5}{2003}+\frac{x+7}{2001}\)

\(\Leftrightarrow\frac{x-1}{2009}+1+\frac{x-2}{2010}+1=\frac{x+5}{2003}+1+\frac{x+7}{2001}+1\)

\(\Leftrightarrow\frac{x-1}{2009}+\frac{2009}{2009}+\frac{x-2}{2010}+\frac{2010}{2010}=\frac{x+5}{2003}+\frac{2003}{2003}+\frac{x+7}{2001}+\frac{2001}{2001}\)

\(\Leftrightarrow\frac{x+2008}{2009}+\frac{x+2008}{2010}=\frac{x+2008}{2003}+\frac{x+2008}{2001}\)

\(\Leftrightarrow\frac{x+2008}{2009}+\frac{x+2008}{2010}-\frac{x+2008}{2003}-\frac{x+2008}{2001}=0\)

\(\Leftrightarrow\left(x+2008\right)\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2003}-\frac{1}{2001}\right)=0\)

\(\Leftrightarrow x+2008=0\Leftrightarrow x=-2008\)

Answer 2

Cộng mỗi p/s thêm 1 bạn nhé :)

 

Answer 3

+1 vào mỗi phân thức ...................

Answer 4

(x-1/2009+1)+(x-2/2010+1)-(x+5/2003+1)-(x+7/2001)