1085 Вправа 1085Алгебра 8 клас Бевз 2025 · § 21 · стор. 204

1085. Розклади на множники тричлен. а) 0,6a² + a − 1,6; б) c² − √2c − 4; в) 1,5y² − 3y + 4/3.

а) ${\displaystyle 0,6a^2+a-1,6=}$${\displaystyle 0,2(3a^2+5a-8)}$;

${\displaystyle 3a^2+5a-8}$: ${\displaystyle D=}$${\displaystyle 5^2-4\cdot 3\cdot (-8)=}$${\displaystyle 121}$;

${\displaystyle a_{1,2}=}$${\displaystyle \frac{-5\pm 11}{6}}$;

${\displaystyle a_1=}$${\displaystyle 1}$;

${\displaystyle a_2=}$${\displaystyle -\frac{8}{3}}$;

${\displaystyle 3a^2+5a-8=}$${\displaystyle 3(a-1)(a+\frac{8}{3})=}$${\displaystyle \left(a-1\right)\left(3a+8\right)}$;

${\displaystyle 0,6a^2+a-1,6=}$${\displaystyle 0,2\left(a-1\right)\left(3a+8\right)}$;

б) ${\displaystyle c^2-\sqrt{2}c-4}$: ${\displaystyle D=}$${\displaystyle 2+16=}$${\displaystyle 18}$;

${\displaystyle \sqrt{18}=}$${\displaystyle 3\sqrt{2}}$;

${\displaystyle c_{1,2}=}$${\displaystyle \frac{\sqrt{2}\pm 3\sqrt{2}}{2}}$;

${\displaystyle c_1=}$${\displaystyle 2\sqrt{2}}$;

${\displaystyle c_2=}$${\displaystyle -\sqrt{2}}$;

${\displaystyle c^2-\sqrt{2}c-4=}$${\displaystyle \left(c-2\sqrt{2}\right)\left(c+\sqrt{2}\right)}$;

в) ${\displaystyle 1,5y^2-3y+\frac{4}{3}}$: ${\displaystyle D=}$${\displaystyle 9-4\cdot 1,5\cdot \frac{4}{3}=}$${\displaystyle 9-8=}$${\displaystyle 1}$;

${\displaystyle y_{1,2}=}$${\displaystyle \frac{3\pm 1}{3}}$;

${\displaystyle y_1=}$${\displaystyle \frac{4}{3}}$;

${\displaystyle y_2=}$${\displaystyle \frac{2}{3}}$;

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

=${\displaystyle \frac{3}{2}\cdot \frac{3y-4}{3}\cdot \frac{3y-2}{3}=}$${\displaystyle \frac{1}{6}\left(3y-4\right)\left(3y-2\right)}$.