g) Ta có: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x-3}=1\)
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+3}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(3x^2-9x-x+3-\left(2x^2-2x+5x-5\right)-x^2+4x-3=0\)
\(\Leftrightarrow2x^2-6x-2x^2-3x+5=0\)
\(\Leftrightarrow-9x=-5\)
hay \(x=\dfrac{5}{9}\)