$a,x(x-1)-x²+2x=5$
$⇔x²-x-x²+2x=5$
$⇔x=5$
$b,2x²-2x=(x-1)²$
$⇔2x²-2x=x²-2x+1$
$⇔2x²-x²-2x+2x=1$
$⇔x²=1$
$⇔x=±√1$
$⇔x=±1$
$c,(x+3)(x²-3x+9)-x(x-2)²=19$
$⇔(x³+3³)-x(x²-2.x.2+2²)=19$
$⇔x³+27-x(x²-4x+4)=19$
$⇔x³+27-x³+4x²-4x=19$
$⇔4x²-4x=19-27$
$⇔4x²-4x=-8$
$⇔4x²-4x+8=0$
$⇔4(x²-x+2)=0$
$⇔4=0$ (loại) hoặc $x²-x+2=0$
$⇔x²-x=-2$ (loại)
Vậy pt c vô nghiệm