Question

Как можно решить уравнение X^2+1/x^2=x-1/x+2

Answer 1

Ответ:

$x^2+\dfrac{1}{x^2}=x-\dfrac{1}{x}+2\ \ ,\ \ \ \ ODZ:\ x\ne 0\ ,\\\\\\t=x-\dfrac{1}{x}\ \ ,\ \ t^2=x^2-2+\dfrac{1}{x^2}=\Big(x^2+\dfrac{1}{x^2}\Big)-2\ \ \to \ \ x^2+\dfrac{1}{x^2}=t^2+2\ ,\\\\\\t^2+2=t+2\ \ ,\ \ t^2=t\ \ ,\ \ t^2-t=0\ \ ,\ \ t(t-1)=0\ ,\\\\t_1=0\ \ ,\ \ t_2=1\\\\a)\ \ x-\dfrac{1}{x}=0\ \ ,\ \ \dfrac{x^2-1}{x}=0\ \ ,\ \ \dfrac{(x-1)(x+1)}{x}=0\ \ ,\ \ x_1=-1\ ,\ x_2=1\\\\b)\ \ x-\dfrac{1}{x}=1\ \ ,\ \ \dfrac{x^2-1}{x}=1\ \ ,\ \ \dfrac{x^2-1-x}{x}=0\ \ ,\ \ x^2-x-1=0\ ,$

$D=1+4=5\ \ ,\ \ x_1=\dfrac{1-\sqrt5}{2}\ \ ,\ \ x_2=\dfrac{1+\sqrt5}{2}\\\\\\Otvet:\ \ x_1=-1\ ,\ x_2=1\ ,\ x_3=\dfrac{1-\sqrt5}{2}\ \ ,\ \ x_4=\dfrac{1+\sqrt5}{2}\ .$