\(\frac{2x}{x-14}-\frac{12x}{2x-28}=0\left(x\ne14\right)\)
\(\Leftrightarrow\frac{4x-12x}{2x-28}=0\Leftrightarrow4x-12x=0\Leftrightarrow-8x=0\Leftrightarrow x=0\) (thoả mãn x khác 14)
Ta có
\(\frac{2x}{x-14}\)-- \(\frac{12x}{2x-28}\)=0
<=>\(\frac{4x}{2x-28}\)=\(\frac{12x}{2x-28}\)
<=>4x=12x
<=>x=0
Vậy phương trình có x=0