841° Вправа 841Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 254

841. Розв’яжіть рівняння:

1) ${\displaystyle \frac{3x}{x-2}=}$${\displaystyle \frac{10-x^2}{x-2}}$;

${\displaystyle 3x=}$${\displaystyle 10-x^2}$, ${\displaystyle x\ne 2}$;

${\displaystyle x^2+3x-10=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 3^2-4\cdot 1\cdot (-10)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3-7}{2}=}$${\displaystyle -5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3+7}{2}=}$${\displaystyle 2}$ — не задовольняє умову ${\displaystyle x\ne 2}$;

2) ${\displaystyle \frac{x^2+20}{x-10}=}$${\displaystyle \frac{-12x}{10-x}}$;

${\displaystyle \frac{x^2+20}{x-10}=}$${\displaystyle \frac{12x}{x-10}}$;

${\displaystyle x^2+20=}$${\displaystyle 12x}$, ${\displaystyle x\ne 10}$;

${\displaystyle x^2-12x+20=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-12)}^2-4\cdot 1\cdot 20=}$${\displaystyle 144-80=}$${\displaystyle 64}$;

${\displaystyle x_1=}$${\displaystyle \frac{12-8}{2}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{12+8}{2}=}$${\displaystyle 10}$ — не задовольняє умову ${\displaystyle x\ne 10}$.