853 Вправа 853Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 255

853. Розв’яжіть рівняння:

1) ${\displaystyle {(x^2+1)}^2-12(x^2+1)=}$${\displaystyle -35}$;

${\displaystyle t=}$${\displaystyle x^2+1}$, ${\displaystyle t\ge 1}$;

${\displaystyle t^2-12t+35=}$${\displaystyle 0}$;

${\displaystyle t_1+t_2=}$${\displaystyle 12}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle 35}$;

${\displaystyle t_1=}$${\displaystyle 5}$; ${\displaystyle t_2=}$${\displaystyle 7}$;

${\displaystyle x^2+1=}$${\displaystyle 5}$ або ${\displaystyle x^2+1=}$${\displaystyle 7}$;

${\displaystyle x^2=}$${\displaystyle 4}$ або ${\displaystyle x^2=}$${\displaystyle 6}$;

${\displaystyle x_{1,2}=}$${\displaystyle \pm 2}$; ${\displaystyle x_{3,4}=}$${\displaystyle \pm \sqrt{6}}$;

2) ${\displaystyle 5{\left(\frac{x+12}{x}\right)}^2+3\left(\frac{x+12}{x}\right)-2=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle \frac{x+12}{x}}$, ${\displaystyle x\ne 0}$;

${\displaystyle 5t^2+3t-2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 3^2-4\cdot 5\cdot (-2)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle t_1=}$${\displaystyle \frac{-3+7}{10}=}$${\displaystyle 0,4}$; ${\displaystyle t_2=}$${\displaystyle \frac{-3-7}{10}=}$${\displaystyle -1}$;

${\displaystyle \frac{x+12}{x}=}$${\displaystyle 0,4}$; ${\displaystyle x+12=}$${\displaystyle 0,4x}$; ${\displaystyle 0,6x=}$${\displaystyle -12}$; ${\displaystyle x=}$${\displaystyle -20}$;

${\displaystyle \frac{x+12}{x}=}$${\displaystyle -1}$; ${\displaystyle x+12=}$${\displaystyle -x}$; ${\displaystyle 2x=}$${\displaystyle -12}$; ${\displaystyle x=}$${\displaystyle -6}$;

${\displaystyle x_1=}$${\displaystyle -20}$; ${\displaystyle x_2=}$${\displaystyle -6}$.