Question

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

Answer 1

ответ:-4;-2-√3;0;-2+√3;

Объяснение:

Answer 2

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

$\displaystyle \tt \bold{x_1=0}\\\\\\ \displaystyle \tt x+4=0\\\displaystyle \tt \bold{x_2=-4}\\\\\\ \displaystyle \tt x^2+4x+1=0\\\displaystyle \tt D=4^2-4\cdot1\cdot1=16-4=12\\\displaystyle \tt \sqrt{D}=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}\\\\ \displaystyle \tt x_3=\frac{-4+2\sqrt{3}}{2}=\frac{-2(2-\sqrt{3})}{2}=-2+\sqrt{3}\\\\ \displaystyle \tt x_4=\frac{-4-2\sqrt{3}}{2}=\frac{-2(2+\sqrt{3})}{2}=-2-\sqrt{3}$

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