\(ĐKXĐ:x\ne-1;x\ne\dfrac{1}{2}\)
Ta có : \(\dfrac{6x+5}{3x+3}=\dfrac{5-4x}{1-2x}\)
\(\Leftrightarrow\left(6x+5\right)\left(1-2x\right)=\left(5-4x\right)\left(3x+3\right)\)
\(\Leftrightarrow6x-12x^2+5-10x=15x+15-12x^2-12x\)
\(\Leftrightarrow5-4x-12x^2-15-3x+12x^2=0\)
\(\Leftrightarrow-10-7x=0\)
\(\Rightarrow x=\dfrac{-10}{7}\)