\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\) ( ĐKXĐ : \(x\ne\frac{3}{5};x\ne\frac{1}{5}\) )
\(\Leftrightarrow\frac{-4}{\left(5x-3\right)\left(5x-1\right)}=\frac{3\left(5x-3\right)}{\left(5x-3\right)\left(5x-1\right)}-\frac{2\left(5x-1\right)}{\left(5x-3\right)\left(5x-1\right)}\)
\(\Leftrightarrow-4=-15x-9-10x+2\)
\(\Leftrightarrow5x=3\)
\(\Leftrightarrow x=\frac{3}{5}\) ( loại )
Vậy phương trình trên vô nghiệm
