997 Вправа 997Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

997. Розв’яжи рівняння, використовуючи метод заміни змінної. а) (2x − 1)⁴ − 10(2x − 1)² + 9 = 0; б) (x² + 2)² − (x² + 2) − 6 = 0; в) (x/(x − 2))² − x/(x − 2) = 42; г) ((√x + 1)/3)² + (√x + 1)/3 = 2; ґ) 4x² − 12x + 2(2x − 3) − 6 = 0; д) x² + 7 = 3√(x² + 7) + 4.

а) ${\displaystyle {(2x-1)}^4-10{(2x-1)}^2+9=}$${\displaystyle 0}$;

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

${\displaystyle t^2-10t+9=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -10,c=}$${\displaystyle 9}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-10)}^2-4\cdot 1\cdot 9=}$${\displaystyle 100-36=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{10\pm \sqrt{64}}{2}=}$${\displaystyle \frac{10\pm 8}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{10+8}{2}=}$${\displaystyle 9}$;

${\displaystyle t_2=}$${\displaystyle \frac{10-8}{2}=}$${\displaystyle 1}$;

${\displaystyle {(2x-1)}^2=}$${\displaystyle 9}$;

${\displaystyle 2x-1=}$${\displaystyle \pm 3}$;

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

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

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

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

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

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

б) ${\displaystyle {(x^2+2)}^2-(x^2+2)-6=}$${\displaystyle 0}$;

заміна ${\displaystyle x^2+2=}$${\displaystyle t}$, ${\displaystyle t\ge 2}$:

${\displaystyle t^2-t-6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-6)=}$${\displaystyle 1+24=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle t_1=}$${\displaystyle \frac{1+5}{2}=}$${\displaystyle 3}$;

${\displaystyle t_2=}$${\displaystyle \frac{1-5}{2}=}$${\displaystyle -2}$;

${\displaystyle t=}$${\displaystyle -2}$ — не підходить;

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

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

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

в) ${\displaystyle {\left(\frac{x}{x-2}\right)}^2-\frac{x}{x-2}=}$${\displaystyle 42}$;

заміна ${\displaystyle \frac{x}{x-2}=}$${\displaystyle t}$, ${\displaystyle x\ne 2}$:

${\displaystyle t^2-t=}$${\displaystyle 42}$;

${\displaystyle t^2-t-42=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -42}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-42)=}$${\displaystyle 1+168=}$${\displaystyle 169}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{169}}{2}=}$${\displaystyle \frac{1\pm 13}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{1+13}{2}=}$${\displaystyle 7}$;

${\displaystyle t_2=}$${\displaystyle \frac{1-13}{2}=}$${\displaystyle -6}$;

${\displaystyle \frac{x}{x-2}=}$${\displaystyle 7}$;

${\displaystyle x=}$${\displaystyle 7x-14}$;

${\displaystyle 6x=}$${\displaystyle 14}$;

${\displaystyle x=}$${\displaystyle \frac{7}{3}}$;

${\displaystyle \frac{x}{x-2}=}$${\displaystyle -6}$;

${\displaystyle x=}$${\displaystyle -6x+12}$;

${\displaystyle 7x=}$${\displaystyle 12}$;

${\displaystyle x=}$${\displaystyle \frac{12}{7}}$;

г) ${\displaystyle {\left(\frac{\sqrt{x}+1}{3}\right)}^2+\frac{\sqrt{x}+1}{3}=}$${\displaystyle 2}$;

заміна ${\displaystyle \frac{\sqrt{x}+1}{3}=}$${\displaystyle t}$, ${\displaystyle t>0}$:

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

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

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

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

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{9}}{2}=}$${\displaystyle \frac{-1\pm 3}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-1+3}{2}=}$${\displaystyle 1}$;

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

${\displaystyle t=}$${\displaystyle -2<0}$ — не підходить;

${\displaystyle \frac{\sqrt{x}+1}{3}=}$${\displaystyle 1}$;

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

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

ґ) ${\displaystyle 4x^2-12x+2(2x-3)-6=}$${\displaystyle 0}$;

оскільки ${\displaystyle 4x^2-12x=}$${\displaystyle {(2x-3)}^2-9}$, то

${\displaystyle {(2x-3)}^2-9+2(2x-3)-6=}$${\displaystyle 0}$;

заміна ${\displaystyle 2x-3=}$${\displaystyle t}$:

${\displaystyle t^2+2t-15=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 2,c=}$${\displaystyle -15}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 1\cdot (-15)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{64}}{2}=}$${\displaystyle \frac{-2\pm 8}{2}}$;

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

${\displaystyle t_2=}$${\displaystyle \frac{-2-8}{2}=}$${\displaystyle -5}$;

${\displaystyle 2x-3=}$${\displaystyle 3}$;

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

${\displaystyle 2x-3=}$${\displaystyle -5}$;

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

д) ${\displaystyle x^2+7=}$${\displaystyle 3\sqrt{x^2+7}+4}$;

заміна ${\displaystyle \sqrt{x^2+7}=}$${\displaystyle t}$, ${\displaystyle t>0}$:

${\displaystyle t^2=}$${\displaystyle 3t+4}$;

${\displaystyle t^2-3t-4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -3,c=}$${\displaystyle -4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{25}}{2}=}$${\displaystyle \frac{3\pm 5}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{3+5}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{3-5}{2}=}$${\displaystyle -1}$;

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

${\displaystyle \sqrt{x^2+7}=}$${\displaystyle 4}$;

${\displaystyle x^2+7=}$${\displaystyle 16}$;

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

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