Question

Cm\(\frac{1}{2x-2006}\)+\(\frac{1}{3-2007x}+\frac{1}{2006x+2005}=\frac{1}{x+2}\)

Answer 1

Giải phương trình chứ chứng minh cái gì

\(\frac{1}{2x-2006}+\frac{1}{3-2007x}+\frac{1}{2006x+2005}=\frac{1}{x+2}\)

\(\Leftrightarrow\left(\frac{1}{2x-2006}-\frac{1}{x+2}\right)+\left(\frac{1}{3-2007x}+\frac{1}{2006x+2005}\right)=0\)

\(\Leftrightarrow\frac{x-2008}{\left(2x-2006\right)\left(x+2\right)}+\frac{x-2008}{\left(3-2007x\right)\left(2006x-2005\right)}=0\)

\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{\left(2x-2006\right)\left(x+2\right)}+\frac{1}{\left(3-2007x\right)\left(2006x-2005\right)}\right)=0\)

\(\Leftrightarrow\left(x-2008\right)\left(2008x-1\right)\left(2005x+2003\right)=0\)

\(\Leftrightarrow x=2008;x=\frac{1}{2008};x=-\frac{2003}{2005}\)