$1)x²-4x-y²+4$
$=(x²-4x+4)-y²$
$=(x-2)²-y²$
$=(x-2-y)(x-2+y)$
$2) 3x²-7x+2$
$=3x²-6x-x+2$
$=3x(x-2)-(x-2)$
$=(3x-1)(x-2)$
$3) (x+1)³-3x(x-4)+15(1-x)=17$
$⇔x³+3x²+3x+1-3x²+12x+15-15x=17$
$⇔x³+16=17$
$⇔x³=17-16$
$⇔x³=1$
$⇔x=1$
$4) (2x-1)²=(x+2)²$
$⇔(2x)²-2.2x.1+1²=x²+2.x.2+2²$
$⇔4x²-4x+1-x²-4x-4=0$
$⇔3x²-8x-3=0$
$⇔3x²-9x+x-3=0$
$⇔3x(x-3)+(x-3)=0$
$⇔(3x+1)(x-3)=0$
$⇔3x+1=0$ hoặc $x-3=0$
$⇔3x=-1$ hoặc $x=3$
$⇔x=-1/3$ hoặc $x=3$