855 Вправа 855Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

855. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x+4}{x-3}+\frac{4}{x}=}$${\displaystyle \frac{7}{x-3}}$; ОДЗ: ${\displaystyle x\ne 0}$, ${\displaystyle x\ne 3}$;

${\displaystyle \frac{{x+4}^{\mathrm{\setminus }x}}{x-3}+\frac{4^{\mathrm{\setminus }x-3}}{x}-\frac{7^{\mathrm{\setminus }x}}{x-3}=}$${\displaystyle 0}$;

${\displaystyle \frac{x(x+4)+4(x-3)-7x}{x(x-3)}=}$${\displaystyle 0}$;

${\displaystyle \frac{x^2+4x+4x-12-7x}{x(x-3)}=}$${\displaystyle 0}$;

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

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

${\displaystyle x_1+x_2=}$${\displaystyle -1}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -12}$;

${\displaystyle x_1=}$${\displaystyle -4}$; ${\displaystyle x_2=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle 3}$ — не задовольняє умову ${\displaystyle x\ne 3}$;

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

2) ${\displaystyle \frac{x-2}{x+2}+\frac{x+2}{x-2}=}$${\displaystyle \frac{5}{2}}$; ОДЗ: ${\displaystyle x\ne -2}$, ${\displaystyle x\ne 2}$;

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

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

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

${\displaystyle \frac{2x^2-8x+8+2x^2+8x+8-5x^2+20}{2(x^2-4)}=}$${\displaystyle 0}$;

${\displaystyle \frac{-x^2+36}{2(x^2-4)}=}$${\displaystyle 0}$;

${\displaystyle -x^2+36=}$${\displaystyle 0}$;

${\displaystyle x^2=}$${\displaystyle 36}$;

${\displaystyle x_1=}$${\displaystyle 6}$; ${\displaystyle x_2=}$${\displaystyle -6}$.