Question

Решите урав (x-2)(x^2+2x+1)=4x(x+1)

Answer 1

$(x-2)(x^2+2x+1)=4x(x+1) \\ (x-2)(x^2+2x+1)-4x(x+1)=0 \\ (x-2)(x+1)^2-4x(x+1)=0 \\ (x+1)((x-2)(x+1)-4x)=0 \\ (x+1)(x^2-x-2-4x)=0 \\ (x+1)(x^2-5x-2)=0 \\ \\ x+1=0 \\ x=-1 \\ \\ x^2-5x-2=0 \\ D=25-4*(-2)=33 \\ x= \frac{5б \sqrt{33} }{2}$

Ответ: $-1; \frac{5б \sqrt{33} }{2}$

Answer 2

⇒х²+2х+1=0; Д=4-4=0; х=-2/2=-1; (х-2)(х+1)(х+1)=4х(х+1); (х-2)(х+1)=4х; х²-2х+х-2=4х; х²-5х-2=0; Д=33; х₁=(5-√33)/2; х₂=(5+√33)/2