`a ) (x - 1)( x + 5 ) = 0`
`⇔` \(\left[ \begin{array}{l}x-1=0\\x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-1\\x=-5\end{array} \right.\)
`\text{Vậy S = { -1 ; -5}}`
`b ) (x+2)/(x-2) - 12/(x^2 - 4) = (x-2)/(x+2)` `ĐKXĐ : x` $\neq$ `2 ;-2`
`⇔ ((x+2)^2)/((x-2)(x+2)) - 12/((x-2)(x+2)) = ((x-2)^2)/((x+2)(x-2))`
`⇒ ( x + 2 )^2 - 12 = ( x - 2 )^2`
`⇔ x^2 + 4x + 4 - 12 = x^2 - 4x + 4`
`⇔ x^2 + 4x + 4x - x^2 = 4 + 12 - 4`
`⇔ 8x =12`
`⇔ x = 3/2`
`\text{Vậy S =}` `{3/2}`
`c ) |3x - 5| = 1 ⇔ 3x - 5 = ±1 `
`⇔ ` \(\left[ \begin{array}{l}3x=6\\3x=4\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=\frac{4}{3}\end{array} \right.\)
`\text{Vậy S =}` `{2;4/3}`
