1388 Вправа 1388Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1388. Розв’яжи рівняння. а) x⁶ − 9x³ + 8 = 0; б) z⁶ − 19z³ − 216 = 0.

а) ${\displaystyle x^6-9x^3+8=}$${\displaystyle 0}$;

заміна ${\displaystyle x^3=}$${\displaystyle t}$:

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -9,c=}$${\displaystyle 8}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-9)}^2-4\cdot 1\cdot 8=}$${\displaystyle 81-32=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{9\pm \sqrt{49}}{2}=}$${\displaystyle \frac{9\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{9+7}{2}=}$${\displaystyle 8}$;

${\displaystyle t_2=}$${\displaystyle \frac{9-7}{2}=}$${\displaystyle 1}$;

${\displaystyle x^3=}$${\displaystyle 8}$;

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

${\displaystyle x^3=}$${\displaystyle 1}$;

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

б) ${\displaystyle z^6-19z^3-216=}$${\displaystyle 0}$;

заміна ${\displaystyle z^3=}$${\displaystyle t}$:

${\displaystyle t^2-19t-216=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -19,c=}$${\displaystyle -216}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-19)}^2-4\cdot 1\cdot (-216)=}$${\displaystyle 361+864=}$${\displaystyle 1225}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{19\pm \sqrt{1225}}{2}=}$${\displaystyle \frac{19\pm 35}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{19+35}{2}=}$${\displaystyle 27}$;

${\displaystyle t_2=}$${\displaystyle \frac{19-35}{2}=}$${\displaystyle -8}$;

${\displaystyle z^3=}$${\displaystyle 27}$;

${\displaystyle z=}$${\displaystyle 3}$;

${\displaystyle z^3=}$${\displaystyle -8}$;

${\displaystyle z=}$${\displaystyle -2}$.