376 Вправа 376Алгебра 8 клас Бевз 2025 · § 8 · стор. 77

376. Користуючись властивістю пропорції, розв’яжи рівняння.

а) ${\displaystyle \frac{0,5}{18-x}=}$${\displaystyle \frac{1}{4}}$;

допустимі ${\displaystyle x\ne 18}$;

${\displaystyle 0,5\cdot 4=}$${\displaystyle 18-x}$;

${\displaystyle 2=}$${\displaystyle 18-x}$;

${\displaystyle x=}$${\displaystyle 18-2}$;

${\displaystyle x=}$${\displaystyle 16}$;

б) ${\displaystyle \frac{5x}{7x-12}=}$${\displaystyle \frac{1}{0,2}}$;

допустимі ${\displaystyle x\ne \frac{12}{7}}$;

${\displaystyle 0,2\cdot 5x=}$${\displaystyle 7x-12}$;

${\displaystyle x=}$${\displaystyle 7x-12}$;

${\displaystyle x-7x=}$${\displaystyle -12}$;

${\displaystyle -6x=}$${\displaystyle -12}$;

${\displaystyle x=}$${\displaystyle \frac{-12}{-6}}$;

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

в) ${\displaystyle \frac{2}{x-1}=}$${\displaystyle \frac{5}{x}}$;

допустимі ${\displaystyle x\ne 0,x\ne 1}$;

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

${\displaystyle 2x=}$${\displaystyle 5x-5}$;

${\displaystyle 2x-5x=}$${\displaystyle -5}$;

${\displaystyle -3x=}$${\displaystyle -5}$;

${\displaystyle x=}$${\displaystyle \frac{-5}{-3}}$;

${\displaystyle x=}$${\displaystyle \frac{5}{3}=}$${\displaystyle 1\frac{2}{3}}$;

г) ${\displaystyle \frac{4}{x-1}=}$${\displaystyle \frac{x+1}{6}}$;

допустимі ${\displaystyle x\ne 1}$;

${\displaystyle (x-1)(x+1)=}$${\displaystyle 4\cdot 6}$;

${\displaystyle x^2-1=}$${\displaystyle 24}$;

${\displaystyle x^2=}$${\displaystyle 24+1}$;

${\displaystyle x^2=}$${\displaystyle 25}$;

${\displaystyle x=}$${\displaystyle \pm 5}$;

ґ) ${\displaystyle \frac{3z-1}{2}=}$${\displaystyle \frac{4}{3z+1}}$;

допустимі ${\displaystyle z\ne -\frac{1}{3}}$;

${\displaystyle (3z-1)(3z+1)=}$${\displaystyle 2\cdot 4}$;

${\displaystyle 9z^2-1=}$${\displaystyle 8}$;

${\displaystyle 9z^2=}$${\displaystyle 8+1}$;

${\displaystyle 9z^2=}$${\displaystyle 9}$;

${\displaystyle z^2=}$${\displaystyle 1}$;

${\displaystyle z=}$${\displaystyle \pm 1}$;

д) ${\displaystyle \frac{2x^2+8}{6x}=}$${\displaystyle \frac{2x}{3}}$;

допустимі ${\displaystyle x\ne 0}$;

${\displaystyle 3(2x^2+8)=}$${\displaystyle 2x\cdot 6x}$;

${\displaystyle 6x^2+24=}$${\displaystyle 12x^2}$;

${\displaystyle 6x^2-12x^2=}$${\displaystyle -24}$;

${\displaystyle -6x^2=}$${\displaystyle -24}$;

${\displaystyle x^2=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle \pm 2}$.