Решите уравнение 1. x(x-1)=02. (x+3)(x-2)=03. (2x-1/3)(3x+3)=0

1.x(x-1)=0
x=0 или x-1=0
x=1
2.(x+3)(x-2)=2
$x^{2} -2x+3x-6=0$
$x^{2} +x-6=0$
$D= b^{2} -4ac= 1^{2} -4*1*(-6)=1+24=25$
$x_{1} = \frac{-1+5}{2} = \frac{4}{2} =2$
$x_{2}= \frac{-1-5}{2}= \frac{-6}{2} =-3$
3.(2x-1/3)(3x+3)
$6 x^{2} +6x-x-1=0$
$6 x^{2} +5x-1=0$
D=25-4*6*(-1)=25+24=49
$x_{1} = \frac{-5+7}{12}= \frac{2}{12} = \frac{1}{6}$
$x_{2} = \frac{-5-7}{12} = \frac{-12}{12} =-1$