864* Вправа 864Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 257

864. Розв’яжіть рівняння:

1) ${\displaystyle x^3-8=}$${\displaystyle (x-2)(x^2+2x+4)}$; ОДЗ: ${\displaystyle x\ne 2}$;

${\displaystyle \frac{5^{\mathrm{\setminus }x-2}}{x^2+2x+4}+\frac{4(x-1)}{(x-2)(x^2+2x+4)}-\frac{1^{\mathrm{\setminus }{x}^2+2x+4}}{x-2}=}$${\displaystyle 0}$;

${\displaystyle \frac{5(x-2)+4(x-1)-(x^2+2x+4)}{(x-2)(x^2+2x+4)}=}$${\displaystyle 0}$;

${\displaystyle 5x-10+4x-4-x^2-2x-4=}$${\displaystyle 0}$;

${\displaystyle -x^2+7x-18=}$${\displaystyle 0}$;

${\displaystyle x^2-7x+18=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-7)}^2-4\cdot 1\cdot 18=}$${\displaystyle 49-72=}$${\displaystyle -23<0}$ — коренів немає;

2) ${\displaystyle x^2=}$${\displaystyle \frac{6}{1-\frac{1}{x}}}$; ОДЗ: ${\displaystyle x\ne 0}$, ${\displaystyle x\ne 1}$;

${\displaystyle x^2=}$${\displaystyle \frac{6}{\frac{x-1}{x}}}$;

${\displaystyle x^2=}$${\displaystyle \frac{6x}{x-1}}$;

${\displaystyle x^2(x-1)=}$${\displaystyle 6x}$;

${\displaystyle x^3-x^2-6x=}$${\displaystyle 0}$;

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

${\displaystyle x=}$${\displaystyle 0}$ — не задовольняє ОДЗ;

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

${\displaystyle x_1+x_2=}$${\displaystyle 1}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -6}$;

${\displaystyle x_1=}$${\displaystyle 3}$; ${\displaystyle x_2=}$${\displaystyle -2}$.