981 Вправа 981Алгебра 8 клас Бевз 2025 · § 19 · стор. 184

981. Використовуючи метод заміни змінної, розв’яжи рівняння. а) x − 7√x + 12 = 0; б) x − 2√x − 3 = 0; в) 3x − 8√x + 5 = 0; г) 2x + 3√x + 1 = 0.

а) ${\displaystyle x-7\sqrt{x}+12=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle t^2-7t+12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -7,c=}$${\displaystyle 12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-7)}^2-4\cdot 1\cdot 12=}$${\displaystyle 49-48=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

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

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

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

${\displaystyle \sqrt{x}=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle 16}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 3}$;

${\displaystyle x=}$${\displaystyle 9}$;

б) ${\displaystyle x-2\sqrt{x}-3=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -2,c=}$${\displaystyle -3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-3)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{2\pm \sqrt{16}}{2}=}$${\displaystyle \frac{2\pm 4}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{2+4}{2}=}$${\displaystyle 3}$;

${\displaystyle t_2=}$${\displaystyle \frac{2-4}{2}=}$${\displaystyle -1}$;

${\displaystyle t=}$${\displaystyle -1<0}$ — не підходить;

${\displaystyle \sqrt{x}=}$${\displaystyle 3}$;

${\displaystyle x=}$${\displaystyle 9}$;

в) ${\displaystyle 3x-8\sqrt{x}+5=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle 3t^2-8t+5=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -8,c=}$${\displaystyle 5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-8)}^2-4\cdot 3\cdot 5=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{8\pm \sqrt{4}}{6}=}$${\displaystyle \frac{8\pm 2}{6}}$;

${\displaystyle t_1=}$${\displaystyle \frac{8+2}{6}=}$${\displaystyle \frac{5}{3}}$;

${\displaystyle t_2=}$${\displaystyle \frac{8-2}{6}=}$${\displaystyle 1}$;

${\displaystyle \sqrt{x}=}$${\displaystyle \frac{5}{3}}$;

${\displaystyle x=}$${\displaystyle \frac{25}{9}}$;

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

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

г) ${\displaystyle 2x+3\sqrt{x}+1=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$, тоді ${\displaystyle x=}$${\displaystyle t^2}$:

${\displaystyle 2t^2+3t+1=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle 3,c=}$${\displaystyle 1}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 2\cdot 1=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{1}}{4}=}$${\displaystyle \frac{-3\pm 1}{4}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-3+1}{4}=}$${\displaystyle -0,5}$;

${\displaystyle t_2=}$${\displaystyle \frac{-3-1}{4}=}$${\displaystyle -1}$;

обидва значення ${\displaystyle t<0}$ — коренів немає.