a, $5x(x – 1) = x – 1$
$⇔5x(x-1)(x-1)=0$
$⇔ (x-1)(5x-1)=0$
$\left[ \begin{array}{l}x-1=0\\5x-1=0\end{array} \right.$
$\left[ \begin{array}{l}x=1\\x=\dfrac{1}{5}\end{array} \right.$
b, $2(x + 5) – x^2 – 5x = 0$
$⇔ 2(x+5)-x(x+5)=0$
$⇔ (x+5)(2-x)=0$
$⇔ \left[ \begin{array}{l}x+5=0\\2-x=0\end{array} \right.$
$⇔ \left[ \begin{array}{l}x=-5\\x=2\end{array} \right.$