Tiếp tuyến $\Delta: y=7x+b\quad(b\ne 1)$
$y=f(x)=\dfrac{2x-3}{5x-4}$
$\to f'(x)=\dfrac{2(5x-4)-5(2x-3)}{(5x-4)^2}$
$=\dfrac{7}{(5x-4)^2}$
Đặt toạ độ tiếp điểm: $M(x_o; f(x_o))$
$\to f'(x_o)=7$
$\to \dfrac{7}{(5x_o-4)^2}=7$
$\to |5x_o-4|=1$
$\to x_o=1$ hoặc $x_o=\dfrac{3}{5}$
$x_o=1\to f(x_o)=-1\to \Delta: y=7(x-1)-1=7x-8$
$x_o=\dfrac{3}{5}\to f(x_o)=\dfrac{9}{5}\to \Delta: y=7\Big(x-\dfrac{3}{5}\Big)+\dfrac{9}{5}=7x-\dfrac{12}{5}$