1392 Вправа 1392Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1392. Розв’яжи рівняння. а) (2x − 1)⁴ − (2x − 1)² − 12 = 0; б) (x + 3)² − 13 + 36/(x + 3)² = 0.

а) ${\displaystyle {(2x-1)}^4-{(2x-1)}^2-12=}$${\displaystyle 0}$;

заміна ${\displaystyle {(2x-1)}^2=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

${\displaystyle t^2-t-12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-12)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle t_1=}$${\displaystyle \frac{1+7}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{1-7}{2}=}$${\displaystyle -3}$;

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

${\displaystyle 2x-1=}$${\displaystyle \pm 2}$;

${\displaystyle x=}$${\displaystyle 1,5}$;

${\displaystyle x=}$${\displaystyle -0,5}$;

${\displaystyle t=}$${\displaystyle -3<0}$ — не підходить;

б) ${\displaystyle {(x+3)}^2-13+\frac{36}{{(x+3)}^2}=}$${\displaystyle 0}$;

заміна ${\displaystyle {(x+3)}^2=}$${\displaystyle t}$, ${\displaystyle t>0}$:

${\displaystyle t-13+\frac{36}{t}=}$${\displaystyle 0\mid \cdot t}$;

${\displaystyle t^2-13t+36=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -13,c=}$${\displaystyle 36}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-13)}^2-4\cdot 1\cdot 36=}$${\displaystyle 169-144=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{13\pm \sqrt{25}}{2}=}$${\displaystyle \frac{13\pm 5}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{13+5}{2}=}$${\displaystyle 9}$;

${\displaystyle t_2=}$${\displaystyle \frac{13-5}{2}=}$${\displaystyle 4}$;

${\displaystyle {(x+3)}^2=}$${\displaystyle 9}$;

${\displaystyle x+3=}$${\displaystyle \pm 3}$;

${\displaystyle x=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle -6}$;

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

${\displaystyle x+3=}$${\displaystyle \pm 2}$;

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

${\displaystyle x=}$${\displaystyle -5}$.