a)$5x(x-1)=x-1$

$⇒5x(x-1)-(x-1)=0$

$⇒(x-1)(5x-1)=0$

$⇒$\(\left[ \begin{array}{l}x=1\\x=\frac{1}{5}\end{array} \right.\)

b)$2(x+5)-x^2-5x=0$

$⇒2(x+5)-x(x+5)=0$

$⇒(x+5)(2-x)=0$

$⇒$\(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\)

a) 5x(x-1) = x - 1

<=> 5x(x-1) - ( x-1) = 0

<=> ( x -1 ) ( 5x - 1 ) = 0

<=> \(\left[ \begin{array}{l}x-1 = 0 \\5x-1=0\end{array} \right.\)

<=> \(\left[ \begin{array}{l}x=1\\5x=1\end{array} \right.\)

<=> \(\left[ \begin{array}{l}x=1\\x=\frac{1}{5}\end{array} \right.\)

Vậy x= 1 hoặc x= $\frac{1}{5}$ 

b) 2(x+5) - $x^{2}$ -5x =0

<=> 2(x+5) -x(x+5) = 0

<=> (x+5) ( 2-x) = 0

<=> \(\left[ \begin{array}{l}x+5=0\\2-x=0\end{array} \right.\) 

<=> \(\left[ \begin{array}{l}x= -5\\x=2\end{array} \right.\)