851 Вправа 851Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 255

851. Розв’яжіть рівняння:

1) ${\displaystyle 3x^4-5x^2+2=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle x^2}$, ${\displaystyle t\ge 0}$;

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 3\cdot 2=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

${\displaystyle t_1=}$${\displaystyle \frac{5-1}{6}=}$${\displaystyle \frac{4}{6}=}$${\displaystyle \frac{2}{3}}$; ${\displaystyle t_2=}$${\displaystyle \frac{5+1}{6}=}$${\displaystyle 1}$;

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

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

2) ${\displaystyle 16x^4-25x^2+9=}$${\displaystyle 0}$;

${\displaystyle t=}$${\displaystyle x^2}$, ${\displaystyle t\ge 0}$;

${\displaystyle 16t^2-25t+9=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-25)}^2-4\cdot 16\cdot 9=}$${\displaystyle 625-576=}$${\displaystyle 49}$;

${\displaystyle t_1=}$${\displaystyle \frac{25-7}{32}=}$${\displaystyle \frac{18}{32}=}$${\displaystyle \frac{9}{16}}$; ${\displaystyle t_2=}$${\displaystyle \frac{25+7}{32}=}$${\displaystyle 1}$;

${\displaystyle x^2=}$${\displaystyle \frac{9}{16}}$ або ${\displaystyle x^2=}$${\displaystyle 1}$;

${\displaystyle x_{1,2}=}$${\displaystyle \pm 0,75}$; ${\displaystyle x_{3,4}=}$${\displaystyle \pm 1}$.