859 Вправа 859Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

859. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x+1}{x+2}-\frac{1}{x-2}=}$${\displaystyle -\frac{4}{x^2-4}}$; ОДЗ: ${\displaystyle x\ne -2}$, ${\displaystyle x\ne 2}$;

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

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

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

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

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

${\displaystyle x_1=}$${\displaystyle 0}$;

${\displaystyle x_2=}$${\displaystyle 2}$ — не задовольняє ОДЗ;

2) ${\displaystyle \frac{3}{x+1}+\frac{x+14}{x^2+x}=}$${\displaystyle 3}$; ОДЗ: ${\displaystyle x\ne 0}$, ${\displaystyle x\ne -1}$;

${\displaystyle \frac{3\cdot x+x+14-3\cdot x(x+1)}{x(x+1)}=}$${\displaystyle 0}$;

${\displaystyle 3x+x+14-3x^2-3x=}$${\displaystyle 0}$;

${\displaystyle -3x^2+x+14=}$${\displaystyle 0}$;

${\displaystyle 3x^2-x-14=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-1)}^2-4\cdot 3\cdot (-14)=}$${\displaystyle 1+168=}$${\displaystyle 169}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{1\pm \sqrt{169}}{2\cdot 3}=}$${\displaystyle \frac{1\pm 13}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+13}{6}=}$${\displaystyle \frac{14}{6}=}$${\displaystyle 2\frac{1}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-13}{6}=}$${\displaystyle -2}$.