Question

розвяжите уравнение
(х²+4х)²+(х+2)²=4​

Answer 1

Ответ:-4;-2-√3;0;-2+√3;

Объяснение:

Answer 2

$displaystyle tt (x^2+4x)^2+(x+2)^2=4\displaystyle tt x^4+8x^3+16x^2+x^2+4x+4=4\displaystyle tt x^4+8x^3+16x^2+x^2+4x=0\displaystyle tt x^4+8x^3+17x^2+4x=0\displaystyle tt x(x^3+8x^3+17x+4)=0\displaystyle tt x(x^3+4x^2+4x^2+16x+x+4)=0\displaystyle tt x(x^2(x+4)+4x(x+4)+1(x+4))=0\displaystyle tt x(x+4)(x^2+4x+1)=0$

$displaystyle tt bold{x_1=0}\\\ displaystyle tt x+4=0\displaystyle tt bold{x_2=-4}\\\ displaystyle tt x^2+4x+1=0\displaystyle tt D=4^2-4cdot1cdot1=16-4=12\displaystyle tt sqrt{D}=sqrt{12}=sqrt{4cdot3}=2sqrt{3}\\ displaystyle tt x_3=frac{-4+2sqrt{3}}{2}=frac{-2(2-sqrt{3})}{2}=-2+sqrt{3}\\ displaystyle tt x_4=frac{-4-2sqrt{3}}{2}=frac{-2(2+sqrt{3})}{2}=-2-sqrt{3}$

ОТВЕТ:

$displaystyle tt x_1=0\displaystyle tt x_2-4\displaystyle tt x_3=-2+sqrt{3}\displaystyle tt x_4=-2-sqrt{3}$