\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)-2=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)-2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\times\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)
\(\Leftrightarrow x=-100\)
Vậy......