861 Вправа 861Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

861. Розв’яжіть рівняння:

1) ${\displaystyle {(x+5)}^4+8{(x+5)}^2-9=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle {(x+5)}^2}$, ${\displaystyle t\ge 0}$;

${\displaystyle t^2+8t-9=}$${\displaystyle 0}$;

${\displaystyle t_1+t_2=}$${\displaystyle -8}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle -9}$;

${\displaystyle t_1=}$${\displaystyle -9}$ — не задовольняє ${\displaystyle t\ge 0}$;

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

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

${\displaystyle x^2+10x+25=}$${\displaystyle 1}$;

${\displaystyle x^2+10x+24=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle -10}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 24}$;

${\displaystyle x_1=}$${\displaystyle -4}$; ${\displaystyle x_2=}$${\displaystyle -6}$;

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

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

${\displaystyle (t+4)t+3=}$${\displaystyle 0}$;

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

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

${\displaystyle t_1=}$${\displaystyle -3}$; ${\displaystyle t_2=}$${\displaystyle -1}$;

${\displaystyle x^2-4x=}$${\displaystyle -3}$ або ${\displaystyle x^2-4x=}$${\displaystyle -1}$;

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

${\displaystyle x_1+x_2=}$${\displaystyle 4}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 3}$;

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

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

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

${\displaystyle x=}$${\displaystyle \frac{4\pm \sqrt{12}}{2}=}$${\displaystyle \frac{4\pm 2\sqrt{3}}{2}=}$${\displaystyle 2\pm \sqrt{3}}$.