$x^2+5x-6=0$
$⇒x^2+6x-x-6=0$
$⇒(x^2+6x)-(x+6)=0$
$⇒x(x+6)-(x+6)=0$
$⇒(x-1)(x+6)=0$
\(⇒\left[ \begin{array}{l}x-1=0⇒x=1\\x+6=0⇒x=-6\end{array} \right.\)
Vậy $S=\{1;-6\}$
$x^2-6x+5=0$
$⇒x^2-5x-x+5=0$
$⇒(x^2-5x)-(x-5)=0$
$⇒x(x-5)-(x-5)=0$
$⇒(x-1)(x-5)=0$
\(⇒\left[ \begin{array}{l}x-1=0⇒x=1\\x-5=0⇒x=5\end{array} \right.\)
Vậy $S=\{1;5\}$