a, 5x( x - 1) = x - 1
⇔ 5x( x - 1 ) - ( x - 1 ) = 0
⇔ ( x -1 ) ( 5x - 1 ) = 0
⇔ \(\left[ \begin{array}{l}x-1=0\\5x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=1/5\end{array} \right.\)
Vậy S = { 1/5; 1}
b, 2( x + 5) - x² - 5x = 0
⇔ 2x + 10 - x² - 5x = 0
⇔ x² + 3x - 10 = 0
⇔ \(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\)