252 Вправа 252Алгебра 8 клас Тарасенкова 2025 · § 7 · стор. 76

252. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x-1}{x^3-1}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x^3-1\ne 0}$;

${\displaystyle (x-1)(x^2+x+1)\ne 0}$;

${\displaystyle x\ne 1}$;

${\displaystyle x-1=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 1}$ — не задовольняє ОДЗ;

коренів немає;

2) ${\displaystyle \frac{x+2}{x^3+8}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x^3+8\ne 0}$;

${\displaystyle (x+2)(x^2-2x+4)\ne 0}$;

${\displaystyle x\ne -2}$;

${\displaystyle x+2=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle -2}$ — не задовольняє ОДЗ;

коренів немає;

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

ОДЗ: ${\displaystyle 3x+6\ne 0}$;

${\displaystyle 3(x+2)\ne 0}$;

${\displaystyle x\ne -2}$;

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

${\displaystyle x-2=}$${\displaystyle 0}$ або ${\displaystyle x+2=}$${\displaystyle 0}$;

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

${\displaystyle x=}$${\displaystyle -2}$ — не задовольняє ОДЗ;

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

4) ${\displaystyle \frac{2x+2}{x^3+1}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x^3+1\ne 0}$;

${\displaystyle (x+1)(x^2-x+1)\ne 0}$;

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

${\displaystyle 2(x+1)=}$${\displaystyle 0}$;

${\displaystyle x+1=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle -1}$ — не задовольняє ОДЗ;

коренів немає;

5) ${\displaystyle \frac{x^3-9x}{15-5x}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle 15-5x\ne 0}$;

${\displaystyle 5(3-x)\ne 0}$;

${\displaystyle x\ne 3}$;

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

${\displaystyle x(x-3)(x+3)=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x-3=}$${\displaystyle 0}$ або ${\displaystyle x+3=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x=}$${\displaystyle 3}$ або ${\displaystyle x=}$${\displaystyle -3}$;

${\displaystyle x=}$${\displaystyle 3}$ — не задовольняє ОДЗ;

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x=}$${\displaystyle -3}$;

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

ОДЗ: ${\displaystyle 4-2x\ne 0}$;

${\displaystyle 2(2-x)\ne 0}$;

${\displaystyle x\ne 2}$;

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

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

${\displaystyle 3x=}$${\displaystyle 0}$ або ${\displaystyle x-2=}$${\displaystyle 0}$ або ${\displaystyle x+2=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x=}$${\displaystyle 2}$ або ${\displaystyle x=}$${\displaystyle -2}$;

${\displaystyle x=}$${\displaystyle 2}$ — не задовольняє ОДЗ;

${\displaystyle x=}$${\displaystyle 0}$ або ${\displaystyle x=}$${\displaystyle -2}$.