Question

Прмогите решить уравнения, пожалуйста. (10 класс)
$3)(x { }^{2} - 3x + 1)(x {}^{2} - 3x + 3) - 3 = 0$
$4)(x {}^{2} + \frac{1}{x {}^{2} } ) + 7(x + \frac{1}{x} ) + 12 = 0$


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Answer 1

Ответ:

$3)\ \ (x^2-3x+1)(x^2-3x+3)-3=0\\\\t=x^2-3x+1\ \ ,\ \ \ t\cdot (t+2)-3=0\ \ ,\ \ t^2+2t-3=0\ \ ,\ \ t_1=-3\ ,\ t_2=1\\\\a)\ \ x^2-3x+1=-3\ ,\ x^2-3x+4=0\ \ ,\ D=9-16=-7<0\ ,\ x\in \varnothing \\\\b)\ \ x^2-3x+1=1\ \ ,\ \ x^2-3x=0\ \ ,\ \ x(x-3)=0\ \ ,\ x_1=0\ ,\ x_2=3\\\\Otvet:\ x_1=0\ ,\ x_2=3\ .$

$4)\ \ \Big(x^2+\dfrac{1}{x^2}\Big)+7\, \Big(x+\dfrac{1}{x}\Big)+12=0\\\\\\t=x+\dfrac{1}{x}\ \ ,\ \ t^2=x^2+2+\dfrac{1}{x^2}= \Big(x^2+\dfrac{1}{x^2}\Big)+2\ \ ,\ \ \ \Big(x^2+\dfrac{1}{x^2}\Big)=t^2-2\ ,\\\\\\t^2-2+7t+12=0\ \ ,\ \ t^2+7t+10=0\ \ ,\ \ t_1=-5\ ,\ t_2=-2\\\\a)\ \ x+\dfrac{1}{x}=-5\ \ ,\ \ \dfrac{x^2+5x+1}{x}=0\ \ ,\ \ x^2+5x+1=0\ ,\ \ D=25-4=21\ ,\\\\x_{1,2}=\dfrac{-5\pm \sqrt{21}}{2}\\\\\\b)\ \ x+\dfrac{1}{x}=-2\ \ ,\ \ \dfrac{x^2+2x+1}{x}=0\ \ ,\ \ (x+1)^2=0\ \ ,\ \ x=-1$

$Otvet:\ \ x_1=\dfrac{-5-\sqrt{21}}{2}\ ,\ x_2=\dfrac{-5+\sqrt{21}}{2}\ ,\ x_3=-1\ .$