860 Вправа 860Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

860. Розв’яжіть рівняння:

1) ${\displaystyle {(x-1)}^4-5{(x-1)}^2+4=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle {(x-1)}^2}$, ${\displaystyle t\ge 0}$;

${\displaystyle t^2-5t+4=}$${\displaystyle 0}$;

${\displaystyle t_1+t_2=}$${\displaystyle 5}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle 4}$;

${\displaystyle t_1=}$${\displaystyle 1}$; ${\displaystyle t_2=}$${\displaystyle 4}$;

${\displaystyle {(x-1)}^2=}$${\displaystyle 1}$ або ${\displaystyle {(x-1)}^2=}$${\displaystyle 4}$;

${\displaystyle x-1=}$${\displaystyle \pm 1}$ або ${\displaystyle x-1=}$${\displaystyle \pm 2}$;

${\displaystyle x_1=}$${\displaystyle 2}$; ${\displaystyle x_2=}$${\displaystyle 0}$; ${\displaystyle x_3=}$${\displaystyle 3}$; ${\displaystyle x_4=}$${\displaystyle -1}$;

2) ${\displaystyle {(x^2+x)}^2-4(x^2+x)+4=}$${\displaystyle 0}$;

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

${\displaystyle t^2-4t+4=}$${\displaystyle 0}$;

${\displaystyle {(t-2)}^2=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle 2}$;

${\displaystyle x^2+x=}$${\displaystyle 2}$;

${\displaystyle x^2+x-2=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle -1}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -2}$;

${\displaystyle x_1=}$${\displaystyle -2}$; ${\displaystyle x_2=}$${\displaystyle 1}$;

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

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

${\displaystyle (t+1)t=}$${\displaystyle 12}$;

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

${\displaystyle t_1+t_2=}$${\displaystyle -1}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle -12}$;

${\displaystyle t_1=}$${\displaystyle -4}$; ${\displaystyle t_2=}$${\displaystyle 3}$;

${\displaystyle x^2+2x=}$${\displaystyle -4}$ або ${\displaystyle x^2+2x=}$${\displaystyle 3}$;

${\displaystyle x^2+2x+4=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 2^2-4\cdot 1\cdot 4=}$${\displaystyle 4-16=}$${\displaystyle -12<0}$ — коренів немає;

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

${\displaystyle x_1+x_2=}$${\displaystyle -2}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -3}$;

${\displaystyle x_1=}$${\displaystyle -3}$; ${\displaystyle x_2=}$${\displaystyle 1}$;

4) ${\displaystyle (x^2+4x+2)(x^2+4x-1)=}$${\displaystyle 10}$;

${\displaystyle x^2+4x=}$${\displaystyle t}$;

${\displaystyle (t+2)(t-1)=}$${\displaystyle 10}$;

${\displaystyle t^2-t+2t-2-10=}$${\displaystyle 0}$;

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

${\displaystyle t_1=}$${\displaystyle -4}$; ${\displaystyle t_2=}$${\displaystyle 3}$;

${\displaystyle x^2+4x=}$${\displaystyle -4}$ або ${\displaystyle x^2+4x=}$${\displaystyle 3}$;

${\displaystyle x^2+4x+4=}$${\displaystyle 0}$;

${\displaystyle {(x+2)}^2=}$${\displaystyle 0}$; ${\displaystyle x=}$${\displaystyle -2}$;

${\displaystyle x^2+4x-3=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 4^2-4\cdot 1\cdot (-3)=}$${\displaystyle 16+12=}$${\displaystyle 28}$;

${\displaystyle x=}$${\displaystyle \frac{-4\pm \sqrt{28}}{2}=}$${\displaystyle \frac{-4\pm 2\sqrt{7}}{2}=}$${\displaystyle -2\pm \sqrt{7}}$.