856 Вправа 856Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

856. Знайдіть усі значення змінної х, за яких:

1) ${\displaystyle \frac{x+1}{5x+1}=}$${\displaystyle \frac{x}{x+2}}$; ОДЗ: ${\displaystyle x\ne -0,2}$, ${\displaystyle x\ne -2}$;

${\displaystyle (x+1)(x+2)=}$${\displaystyle x(5x+1)}$;

${\displaystyle x^2+2x+x+2=}$${\displaystyle 5x^2+x}$;

${\displaystyle x^2+3x+2-5x^2-x=}$${\displaystyle 0}$;

${\displaystyle -4x^2+2x+2=}$${\displaystyle 0}$ | ${\displaystyle :\left(-2\right)}$

${\displaystyle 2x^2-x-1=}$${\displaystyle 0}$;

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

${\displaystyle x_{1,2}=}$${\displaystyle \frac{1\pm \sqrt{9}}{2\cdot 2}=}$${\displaystyle \frac{1\pm 3}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+3}{4}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-3}{4}=}$${\displaystyle \frac{-2}{4}=}$${\displaystyle -0,5}$;

2) ${\displaystyle \frac{2x-2}{x+3}+\frac{x+3}{x-3}=}$${\displaystyle 5}$; ОДЗ: ${\displaystyle x\ne -3}$, ${\displaystyle x\ne 3}$;

${\displaystyle \frac{{2x-2}^{\mathrm{\setminus }x-3}}{x+3}+\frac{{x+3}^{\mathrm{\setminus }x+3}}{x-3}-5^{\mathrm{\setminus }\left(x+3\right)\left(x-3\right)}=}$${\displaystyle 0}$;

${\displaystyle \frac{(2x-2)(x-3)+{(x+3)}^2-5(x+3)(x-3)}{(x+3)(x-3)}=}$${\displaystyle 0}$;

${\displaystyle (2x^2-6x-2x+6)+(x^2+6x+9)-5(x^2-9)=}$${\displaystyle 0}$;

${\displaystyle 2x^2-8x+6+x^2+6x+9-5x^2+45=}$${\displaystyle 0}$;

${\displaystyle -2x^2-2x+60=}$${\displaystyle 0}$ | ${\displaystyle :\left(-2\right)}$

${\displaystyle x^2+x-30=}$${\displaystyle 0}$;

${\displaystyle x_1+x_2=}$${\displaystyle -1}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -30}$;

${\displaystyle x_1=}$${\displaystyle -6}$; ${\displaystyle x_2=}$${\displaystyle 5}$.