Question

решите симметрические уравнения:
1. x⁴+2x³-x²+2x+1
2. x⁴+2x³-x²-2x+1​

Answer 1

Будет (2-4)куб

Объяснение:

Ну я так сделал

Answer 2

$\displaystyle x^4+2x^3-x^2+2x+1=0\\\\(x^4+1)+2(x^3+x)-x^2=0| :x^2\\\\(x^2+\frac{1}{x^2})+2(x+\frac{1}{x})-1=0\\\\x+\frac{1}{x}=t; x^2+2*x*\frac{1}{x}+\frac{1}{x^2}=t^2; x^2+\frac{1}{x^2}=t^2-2 \\\\t^2-2+2t-1=0\\\\t^2+2t-3=0\\\\D=4+12=16\\\\t_{1.2}=\frac{-2 \pm 4}{2}; t_1=-3; t_2=1$

1. t= 1

$\displaystyle x+\frac{1}{x}=1\\\\x^2+1=x\\\\x^2-x+1=0\\\\D=1-4$

корней нет

2. t=-3

$\displaystyle x+\frac{1}{x}=-3\\\\x^2+1=-3x\\\\x^2+3x+1=0\\\\D=9-4=5\\\\ x_{1.2}= \frac{-3 \pm \sqrt{5} }{2}$

ответ х=(-3 ±√5)/2

_______

$\displaystyle x^4+2x^3-x^2-2x+1=0\\\\(x^4+1) +2(x^3-x)-x^2=0|:x^2\\\\(x^2+\frac{1}{x^2})+2(x-\frac{1}{x})-1=0\\\\x-\frac{1}{x}=t; x^2-2*x*\frac{1}{x}+\frac{1}{x^2}=t^2; x^2+\frac{1}{x^2}=t^2+2\\\\t^2+2+2t-1=0\\\\t^2+2t+1=0\\\\D=4-4=0; t=-1$

t=-1

$\displaystyle x-\frac{1}{x}= -1\\\\x^2-1+x=0\\\\D=1+4=5\\\\x_{1.2}=\frac{-1 \pm \sqrt{5}}{2}$

ответ x=(-1±√5)/2