850 Вправа 850Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 255

850. Розв’яжіть рівняння:

1) ${\displaystyle 4x^4-37x^2+9=}$${\displaystyle 0}$;

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

${\displaystyle 4t^2-37t+9=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-37)}^2-4\cdot 4\cdot 9=}$${\displaystyle 1369-144=}$${\displaystyle 1225}$;

${\displaystyle t_1=}$${\displaystyle \frac{37-35}{8}=}$${\displaystyle \frac{2}{8}=}$${\displaystyle \frac{1}{4}}$; ${\displaystyle t_2=}$${\displaystyle \frac{37+35}{8}=}$${\displaystyle \frac{72}{8}=}$${\displaystyle 9}$;

${\displaystyle x^2=}$${\displaystyle \frac{1}{4}}$ або ${\displaystyle x^2=}$${\displaystyle 9}$;

${\displaystyle x_{1,2}=}$${\displaystyle \pm 0,5}$; ${\displaystyle x_{3,4}=}$${\displaystyle \pm 3}$;

2) ${\displaystyle 9x^4-32x^2-16=}$${\displaystyle 0}$;

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

${\displaystyle 9t^2-32t-16=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-32)}^2-4\cdot 9\cdot (-16)=}$${\displaystyle 1024+576=}$${\displaystyle 1600}$;

${\displaystyle t_1=}$${\displaystyle \frac{32+40}{18}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{32-40}{18}=}$${\displaystyle -\frac{4}{9}}$ — не задовольняє ${\displaystyle t\ge 0}$;

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

${\displaystyle x_{1,2}=}$${\displaystyle \pm 2}$;

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

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

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 2=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle t_1=}$${\displaystyle \frac{5+3}{4}=}$${\displaystyle 2}$; ${\displaystyle t_2=}$${\displaystyle \frac{5-3}{4}=}$${\displaystyle 0,5}$;

${\displaystyle x^2=}$${\displaystyle 2}$ або ${\displaystyle x^2=}$${\displaystyle 0,5}$;

${\displaystyle x_{1,2}=}$${\displaystyle \pm \sqrt{2}}$; ${\displaystyle x_{3,4}=}$${\displaystyle \pm \sqrt{0,5}=}$${\displaystyle \pm \frac{\sqrt{2}}{2}}$;

4) ${\displaystyle 16x^4-8x^2+1=}$${\displaystyle 0}$;

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

${\displaystyle 16t^2-8t+1=}$${\displaystyle 0}$;

${\displaystyle {(4t-1)}^2=}$${\displaystyle 0}$;

${\displaystyle 4t=}$${\displaystyle 1}$ | ${\displaystyle :4}$

${\displaystyle t=}$${\displaystyle 0,25}$;

${\displaystyle x^2=}$${\displaystyle 0,25}$;

${\displaystyle x_{1,2}=}$${\displaystyle \pm 0,5}$.