Question

Помогите решить пример

Answer 1

Пошаговое объяснение:

$3x-x^2=\frac{(x^2-3x)^2}{2} \\3x-x^2 = \frac{x^4-6x^3+9x^2}{2} \\6x-2x^2-x^4-6x^3+9x^2\\6x-2x^2-x^4+6x^3-9x^2=0\\6x-11x^2-x^4+6x^3=0\\x*(6-11x-x^3+6x^2)=0\\x*(-x^3+6x^2-11x+6)=0\\x*(-x^3+x^2+5x^2_5x-6x+6)=0\\x*(-x^2*(x-1)+5x*(x-1)-6(x-1))=0\\x*(-(x-1)*(x^2-5x+6))+0\\x*(-(x-1)*(x^2-2x-3x+6))=0\\x*(-(x-1)*(x*(x*(x-2)-3(x-2)))=0\\x*(-(x-1)*(x-2)*(x-3))=0\\-x*(x-1)*(x-2)*(x-3)=0\\x*(x-1)*(x-2)*(x-3)=0\\\\x=0\\x-1=0\\x-2=0\\x-3=0\\\\\\x=0\\x=1\\x=2\\x=3$

Ответ:

$x_{1} =0, x_{2} =1, x_{3} =2, x_{4} = 3$