\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\) (1)
ĐKXĐ : \(x\ne1,-\dfrac{4}{3}\)
\(\left(1\right)\Rightarrow\left(4x+7\right)\left(3x+4\right)=\left(x-1\right)\left(12x+5\right)\)
\(\Leftrightarrow12x^2+16x+21x+28=12x^2+5x-12x-5\)
\(\Leftrightarrow44x+33=0\)
\(\Leftrightarrow x=-\dfrac{33}{44}=-\dfrac{3}{4}\)