Question

Как решить уравнение x^2+4/x^2=x+2/x+2

Answer 1

Ответ:

${x}^{2} + \frac{4}{ {x}^{2} } = x + \frac{2}{x} + 2$

$\frac{ {x}^{4} + 4 }{ {x}^{2} } = \frac{ {x}^{2} + 2x + 2 }{x}$

х≠0

$x( {x}^{4} + 4) = {x}^{2} ( {x}^{2} + 2x + 2)$

${x}^{5} + 4x - {x}^{4} - 2 {x}^{3} - 2 {x}^{2} = 0$

$x(x - 2)(x - 1)(x + 2x + 2) = 0$

x1≠0

x2=2

x3=1

x+2x+2=0

D<0

x2=2

x3=1

Answer 2

Ответ:

$x^2+\dfrac{4}{x^2}=x+\dfrac{2}{x}+2\ \ ,\ \ \ \ ODZ:\ x\ne 0\ ,\\\\\\t=x+\dfrac{2}{x}\ \ ,\ \ t^2=x^2+4+\dfrac{4}{x^2}=\Big(x^2+\dfrac{4}{x^2}\Big)+4\ \ \to \ \ x^2+\dfrac{1}{x^2}=t^2-4\ ,\\\\\\t^2-4=t+2\ \ ,\ \ t^2-t-6=0\ \ ,\ \ t_1=-2\ \ ,\ \ t_2=3\ \ (teorema\ Vieta)\\\\a)\ \ x+\dfrac{2}{x}=-2\ \ ,\ \ \dfrac{x^2+2}{x}+2=0\ \ ,\ \ \dfrac{x^2+2x+2}{x}=0\\\\\\x^2+2x+2=0\ \ \ \ ,\ \ D/4=1-2=-1<0\ \ \Rightarrow\ \ x\in \varnothing$

$b)\ \ x+\dfrac{2}{x}=3\ \ ,\ \ \dfrac{x^2+2}{x}-3=0\ \ ,\ \ \dfrac{x^2-3x+2}{x}=0\ \ ,\\\\\\x^2-3x+2=0\ ,\ \ x_1=1\ ,\ x_2=2\ \ (teorema\ Vieta)\\\\\\Otvet:\ \ x_1=1\ ,\ x_2=2\ .$