48 Вправа 48 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

48. Розв’яжіть рівняння:

1) ${\displaystyle x^4-26x^2+25=}$${\displaystyle 0}$;

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

${\displaystyle t^2-26t+25=}$${\displaystyle 0}$;

${\displaystyle t_1+t_2=}$${\displaystyle 26}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle 25}$;

${\displaystyle t_1=}$${\displaystyle 1}$; ${\displaystyle t_2=}$${\displaystyle 25}$;

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

${\displaystyle x=}$${\displaystyle \pm \sqrt{1}}$ або ${\displaystyle x=}$${\displaystyle \pm \sqrt{25}}$;

${\displaystyle x=}$${\displaystyle \pm 1}$ або ${\displaystyle x=}$${\displaystyle \pm 5}$;

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

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

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

${\displaystyle t_1+t_2=}$${\displaystyle -1}$; ${\displaystyle t_1\cdot t_2=}$${\displaystyle -2}$;

${\displaystyle t_1=}$${\displaystyle -2}$ — не задовольняє умову ${\displaystyle t\ge 0}$; ${\displaystyle t_2=}$${\displaystyle 1}$;

${\displaystyle x^2=}$${\displaystyle 1}$;

${\displaystyle x=}$${\displaystyle \pm \sqrt{1}}$;

${\displaystyle x=}$${\displaystyle \pm 1}$;

3) ${\displaystyle x^4-40x^2+144=}$${\displaystyle 0}$;

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

${\displaystyle t^2-40t+144=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-40)}^2-4\cdot 1\cdot 144=}$${\displaystyle 1600-576=}$${\displaystyle 1024}$;

${\displaystyle t_{1,2}=}$${\displaystyle \frac{40\pm \sqrt{1024}}{2\cdot 1}=}$${\displaystyle \frac{40\pm 32}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{40+32}{2}=}$${\displaystyle 36}$;

${\displaystyle t_2=}$${\displaystyle \frac{40-32}{2}=}$${\displaystyle 4}$;

${\displaystyle x^2=}$${\displaystyle 36}$ або ${\displaystyle x^2=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle \pm \sqrt{36}}$ або ${\displaystyle x=}$${\displaystyle \pm \sqrt{4}}$;

${\displaystyle x=}$${\displaystyle \pm 6}$ або ${\displaystyle x=}$${\displaystyle \pm 2}$;

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

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

${\displaystyle 9t^2+8t-1=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 8^2-4\cdot 9\cdot (-1)=}$${\displaystyle 64+36=}$${\displaystyle 100}$;

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-8\pm \sqrt{100}}{2\cdot 9}=}$${\displaystyle \frac{-8\pm 10}{18}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-8-10}{18}=}$${\displaystyle -1}$ — не задовольняє умову ${\displaystyle t\ge 0}$;

${\displaystyle t_2=}$${\displaystyle \frac{-8+10}{18}=}$${\displaystyle \frac{\stackrel{1}{\cancel{2}}}{\underset{9}{\cancel{18}}}=}$${\displaystyle \frac{1}{9}}$;

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

${\displaystyle x=}$${\displaystyle \pm \sqrt{\frac{1}{9}}}$;

${\displaystyle x=}$${\displaystyle \pm \frac{1}{3}}$.