764 Вправа 764Алгебра 8 клас Тарасенкова 2025 · § 18 · стор. 227

764. Розв’яжіть рівняння:

1) ${\displaystyle \frac{x^2-4}{5}=}$${\displaystyle \frac{x-2}{3}}$ | ${\displaystyle \cdot 15}$

${\displaystyle \frac{(x^2-4)\cdot \stackrel{3}{\cancel{15}}}{\underset{1}{\cancel{5}}}=}$${\displaystyle \frac{(x-2)\cdot \stackrel{5}{\cancel{15}}}{\underset{1}{\cancel{3}}}}$;

${\displaystyle 3(x^2-4)=}$${\displaystyle 5(x-2)}$;

${\displaystyle 3x^2-12=}$${\displaystyle 5x-10}$;

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 3\cdot (-2)=}$${\displaystyle 25+24=}$${\displaystyle 49}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{49}}{2\cdot 3}=}$${\displaystyle \frac{5\pm 7}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+7}{6}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-7}{6}=}$${\displaystyle \frac{-2}{6}=}$${\displaystyle -\frac{1}{3}}$;

2) ${\displaystyle \frac{y(y+1)}{4}-\frac{2}{5}=}$${\displaystyle \frac{3y+2y^2}{20}}$ | ${\displaystyle \cdot 20}$

${\displaystyle \frac{y(y+1)\cdot \stackrel{5}{\cancel{20}}}{\underset{1}{\cancel{4}}}-\frac{2\cdot \stackrel{4}{\cancel{20}}}{\underset{1}{\cancel{5}}}=}$${\displaystyle \frac{(3y+2y^2)\cdot \cancel{20}}{\cancel{20}}}$;

${\displaystyle 5y(y+1)-2\cdot 4=}$${\displaystyle 3y+2y^2}$;

${\displaystyle 5y(y+1)-8=}$${\displaystyle 3y+2y^2}$;

${\displaystyle 5y^2+5y-8-3y-2y^2=}$${\displaystyle 0}$;

${\displaystyle 3y^2+2y-8=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 2^2-4\cdot 3\cdot (-8)=}$${\displaystyle 4+96=}$${\displaystyle 100}$;

${\displaystyle y_{1,2}=}$${\displaystyle \frac{-2\pm \sqrt{100}}{2\cdot 3}=}$${\displaystyle \frac{-2\pm 10}{6}}$;

${\displaystyle y_1=}$${\displaystyle \frac{-2+10}{6}=}$${\displaystyle \frac{8}{6}=}$${\displaystyle 1\frac{1}{3}}$;

${\displaystyle y_2=}$${\displaystyle \frac{-2-10}{6}=}$${\displaystyle -2}$.