Question

решите уравнение x(x-2)(x+1)=xв квадрате (x-1) СРОЧНО!!!

Answer 1

$x(x-2)(x+1)= x^{2} (x-1) \\ x( x^{2} +x-2x-2)=x^{3} - x^{2} \\ x^{3} + x^{2} -2 x^{2} -2x= x^{3} - x^{2} \\ x^{3} - x^{3} + x^{2} -2 x^{2} + x^{2} -2x=0 \\ 2x^{2} -2 x^{2} -2x=0 \\ -2x=0 \\ x=0$

Answer 2

x(x-2)(x+1)=x^2(x-1)
x(x^2+x-2x+2)=x^3-x^2
x^3+x^2-2x^2+2x=x^3-x^2
x^3+x^2-2x^2+2x-x^3+x^2=0
x^2-2x^2+2x+x^2=0
2x^2-2x^2+2x=0
2x=0, |:5
x=0