37 Вправа 37 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 320

37. Знайдіть корені рівняння:

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

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

${\displaystyle D=}$${\displaystyle {(-10)}^2-4\cdot 3\cdot 3=}$${\displaystyle 100-36=}$${\displaystyle 64}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{10\pm \sqrt{64}}{2\cdot 3}=}$${\displaystyle \frac{10\pm 8}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{10+8}{6}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{10-8}{6}=}$${\displaystyle \frac{2}{6}=}$${\displaystyle \frac{1}{3}}$;

2) ${\displaystyle 9y+2=}$${\displaystyle 5y^2}$;

${\displaystyle 5y^2-9y-2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-9)}^2-4\cdot 5\cdot (-2)=}$${\displaystyle 81+40=}$${\displaystyle 121}$;

${\displaystyle y_{1,2}=}$${\displaystyle \frac{9\pm \sqrt{121}}{2\cdot 5}=}$${\displaystyle \frac{9\pm 11}{10}}$;

${\displaystyle y_1=}$${\displaystyle \frac{9+11}{10}=}$${\displaystyle 2}$;

${\displaystyle y_2=}$${\displaystyle \frac{9-11}{10}=}$${\displaystyle \frac{-2}{10}=}$${\displaystyle -0,2}$;

3) ${\displaystyle 9x^2+4=}$${\displaystyle 12x}$;

${\displaystyle 9x^2-12x+4=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-12)}^2-4\cdot 9\cdot 4=}$${\displaystyle 144-144=}$${\displaystyle 0}$;

${\displaystyle x=}$${\displaystyle \frac{12}{18}=}$${\displaystyle \frac{2}{3}}$;

4) ${\displaystyle 4y^2=}$${\displaystyle 20y-25}$;

${\displaystyle 4y^2-20y+25=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-20)}^2-4\cdot 4\cdot 25=}$${\displaystyle 400-400=}$${\displaystyle 0}$;

${\displaystyle y=}$${\displaystyle \frac{20}{8}=}$${\displaystyle 2,5}$.