$x^2-4x+3>0$
$⇔x^2-x-3x+3>0$
$⇔x(x-1)-3(x-1)>0$
$⇔(x-1)(x-3)>0$
$⇔$\(\left[ \begin{array}{l}\begin{cases}x-1>0\\x-3>0\end{cases}\\\begin{cases}x-1<0\\x-3<0\end{cases}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}\begin{cases}x>1\\x>3\end{cases}\\\begin{cases}x<1\\x<3\end{cases}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x>3\\x<1\end{array} \right.\)