1) $(3x-5)^{2}$ - $(3x+1)^{2}$ = 8
<=> $9x^{2}$ - 30x + 25 - $9x^{2}$ - 6x -1 = 8
<=> -36x = -16
<=> x= = $\frac{4}{9}$
2) 2x.(8x-3)-(4x-3) ²=27
<=> $16x^{2}$ - 6x - $12x^{2}$ + 24x - 9 = 27
<=> $4x^{2}$ + 18x - 36 = 0
<=> (2x-3)(x+6) = 0
TH1: 2x-3 = 0
2x = 3
x= $\frac{3}{2}$
TH2: x+6=0
x=6
4) (2x-3) ²-(2x+1) ²=-3
$4x^{2}$ - 12x + 9 - $4x^{2}$ - 4x - 1 = -3
-16x = -11
x = $\frac{11}{16}$
3) (x+5) ²-x ²=45
$x^{2}$ + 10x +25 - $x^{2}$ = 45
10x = 20
x =2