Question

Решите уравнения методом замены новой переменной​

Answer 1

$1)\ \ 2\, (x^2+\dfrac{1}{x^2})-7\, (x+\dfrac{1}{x})+9=0\ \ ,\ \ \ ODZ:\ x\ne 0\\\\t=x+\dfrac{1}{x}\ \ \to \ \ t^2=x^2+2\cdot x\cdot \dfrac{1}{x}+\dfrac{1}{x^2}=x^2+\dfrac{1}{x^2}+2\ \ ,\ \ x^2+\dfrac{1}{x^2}=t^2-2\\\\\\2(t^2-2)-7t+9=0\ \ ,\ \ 2t^2-7t+5=0\ \ ,\ \ D=9\ ,\\\\t_1=\dfrac{7-3}{4} =1\ ,\ t_2=\dfrac{7+3}{4}=\dfrac{5}{2}\\\\a)\ \ x+\dfrac{1}{x}=1\ \ ,\ \ \dfrac{x^2-x+1}{x}=0\ \ ,\ \ x^2-x+1=0\ \ ,\ \ D=-3$

$b)\ \ x+\dfrac{1}{x}=\dfrac{5}{2}\ \ ,\ \ \dfrac{2x^2-5x+2}{2x}=0\ \ ,\ \ 2x^2-5x+2=0\ \ ,\ \ D=9\ ,\\\\\\x_1=\dfrac{5-3}{4}=\dfrac{1}{2}\ \ ,\ \ x_2=\dfrac{5+3}{4}=2\\\\\\Otvet:\ \ x_1=0,5\ \ ,\ \ x_2=2\ .$

$2)\ \ 6\, (x^2+\dfrac{1}{x^2})+5\, (x+\dfrac{1}{x})-38=0\ \ ,\ x\ne 0\\\\t=x+\dfrac{1}{x}\ \ \to \ \ t^2=x^2+2\cdot x\cdot \dfrac{1}{x}+\dfrac{1}{x^2}=x^2+\dfrac{1}{x^2}+2\ \ ,\ \ x^2+\dfrac{1}{x^2}=t^2-2\\\\\\6(t^2-2)+5t-38=0\ \ ,\ \ 6t^2+5t-50=0\ \ ,\ \ D=1225\ ,\\\\\\t_1=\dfrac{-5-35}{12}=-\dfrac{10}{3}\ ,\ t_2=\dfrac{-5+35}{12}=\dfrac{5}{2}\\\\\\a)\ \ x+\dfrac{1}{x}=-\dfrac{10}{3}\ \ ,\ \ \dfrac{3x^2+10x+3}{3x}=0\ \ ,\ \ 3x^2+10x+3=0\ \ ,\\\\\\D/4=16\ ,\ x_1=\dfrac{-5-4}{3}=-3\ \ ,\ \ x_2=-\dfrac{1}{3}$

$b)\ \ x+\dfrac{1}{x}=\dfrac{5}{2}\ \ ,\ \ \dfrac{2x^2-5x+2}{2x}=0\ \ ,\ \ 2x^2-5x+2=0\ \ ,\ \ D=9\ ,\\\\\\x_1=\dfrac{5-3}{4}=\dfrac{1}{2}\ \ ,\ \ x_2=\dfrac{5+3}{4}=2\\\\\\Otvet:\ \ x=-3\ ,\ x=-\dfrac{1}{3}\ ,\ x=\dfrac{1}{2}\ ,\ x=2\ .$

$3)\ \ (x^2+\dfrac{1}{x^2})+7\, (x-\dfrac{1}{x})+10=0\ \ ,\ \ \ ODZ:\ x\ne 0\\\\t=x-\dfrac{1}{x}\ \ \to \ \ t^2=x^2-2\cdot x\cdot \dfrac{1}{x}+\dfrac{1}{x^2}=x^2+\dfrac{1}{x^2}-2\ \ ,\ \ x^2+\dfrac{1}{x^2}=t^2+2\\\\\\(t^2+2)+7t+10=0\ \ ,\ \ t^2+7t+12=0\ \ ,\ \ D=1\ ,\\\\t_1=-4\ ,\ t_2=-3\\\\a)\ \ x-\dfrac{1}{x}=-4\ \ ,\ \ \dfrac{x^2+4x-1}{x}=0\ \ ,\ \ x^2+4x-1=0\ \ ,\\\\\\D/4=5\ \ ,\ \ \ x_1=-2-\sqrt5\ \ ,\ \ \ x_2=-2+\sqrt5$

$b)\ \ x-\dfrac{1}{x}=-3\ \ ,\ \ \dfrac{x^2+3x-1}{x}=0\ \ ,\ \ x^2+3x-1=0\ \ ,\ \ D=13\ ,\\\\x_1=\dfrac{-3-\sqrt{13}}{2}\ \ ,\ \ x_2=\dfrac{-3+\sqrt{13}}{2}\\\\Otvet:\ \ x=-2-\sqrt5\ ,\ \ x=-2+\sqrt5\ \ ,\ x=\dfrac{-3-\sqrt{13}}{2}\ \ ,\ x=\dfrac{-3+\sqrt{13}}{2}\ .$