800 Вправа 800Алгебра 8 клас Тарасенкова 2025 · § 19 · стор. 236

800. Не обчислюючи коренів рівняння x² − 3√3x − 12 = 0, знайдіть:

${\displaystyle x_1+x_2=}$${\displaystyle 3\sqrt{3}}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -12}$;

1) ${\displaystyle {x_1}^2+{x_2}^2=}$${\displaystyle {(x_1+x_2)}^2-2x_1{x}_2=}$${\displaystyle {\left(3\sqrt{3}\right)}^2-2\cdot (-12)=}$${\displaystyle 27+24=}$${\displaystyle 51}$;

2) ${\displaystyle \frac{1^{\mathrm{\setminus }{x}_2}}{x_1}+\frac{1^{\mathrm{\setminus }{x}_1}}{x_2}=}$${\displaystyle \frac{x_2+x_1}{x_1\cdot x_2}=}$${\displaystyle \frac{3\sqrt{3}}{-12}=}$${\displaystyle -\frac{\stackrel{1}{\cancel{3}}\sqrt{3}}{\underset{4}{\cancel{12}}}=}$${\displaystyle -\frac{\sqrt{3}}{4}}$;

3) ${\displaystyle \frac{{x_2}^{\mathrm{\setminus }{x}_2}}{x_1}+\frac{{x_1}^{\mathrm{\setminus }{x}_1}}{x_2}=}$${\displaystyle \frac{{x_2}^2+{x_1}^2}{x_1\cdot x_2}=}$${\displaystyle \frac{51}{-12}=}$${\displaystyle -\frac{\stackrel{17}{\cancel{51}}}{\underset{4}{\cancel{12}}}=}$${\displaystyle -4,25}$;

4) ${\displaystyle {x_1}^3+{x_2}^3=}$${\displaystyle (x_1+x_2)({x_1}^2-x_1{x}_2+{x_2}^2)=}$${\displaystyle 3\sqrt{3}\cdot (51-(-12))=}$${\displaystyle 3\sqrt{3}\cdot 63=}$${\displaystyle 189\sqrt{3}}$.