4x2 - 1 = (2x + 1)(3x - 5)
<=> 6x2 - 10x + 3x - 5 - 4x2 + 1 = 0
<=> 2x2 - 7x - 4 = 0
<=> 2x2 - 8x + x - 4 = 0
<=> 2x(x - 4) + (x - 4) = 0
<=> (x - 4)(2x + 1) = 0
<=> \(\left[{}\begin{matrix}2x+1=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=4\end{matrix}\right.\)
Vậy S = {- 0,5 : 4}
(2x+1)(2x-1)=(2x+1)(3x-5)
<=>(2x-1)(2x+1)-(2x-1)(3x-5)=0
<=>(2x+1)(2x-1-3x+5)=0
<=>(2x+1)(4-x)=0
=> (2x+1)=0 hoac (4-x)=0
+, 2x+1=0 <=>2x=-1 <=>x=\(\dfrac{-1}{2}\)
+, 4-x=0 <=>x=4
vay x thuộc tap hợp \(\dfrac{-1}{2};4\)