Đáp án:
$\dfrac{x-1}{3x+2}$
Giải thích các bước giải:
$\dfrac{x^2-4x+3}{3x^2-7x-6}$
$=\dfrac{x^2-x-3x+3}{3x^2-9x+2x-6}$
$=\dfrac{(x^2-x)-(3x-3)}{(3x^2-9x)+(2x-6)}$
$=\dfrac{x.(x-1)-3.(x-1)}{3x.(x-3)+2.(x-3)}$
$=\dfrac{(x-1).(x-3)}{(x-3).(3x+2)}$
$=\dfrac{x-1}{3x+2}$