801 Вправа 801Алгебра 8 клас Тарасенкова 2025 · § 19 · стор. 236

801. Не обчислюючи коренів рівняння x² − √2x − 4 = 0, знайдіть:

${\displaystyle x_1+x_2=}$${\displaystyle \sqrt{2}}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle -4}$;

1) ${\displaystyle {x_1}^2+{x_2}^2=}$${\displaystyle {(x_1+x_2)}^2-2x_1{x}_2=}$${\displaystyle {\left(\sqrt{2}\right)}^2-2\cdot (-4)=}$${\displaystyle 2+8=}$${\displaystyle 10}$;

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{\sqrt{2}}{-4}=}$${\displaystyle -\frac{\sqrt{2}}{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{10}{-4}=}$${\displaystyle -\frac{\stackrel{5}{\cancel{10}}}{\underset{2}{\cancel{4}}}=}$${\displaystyle -2,5}$;

4) ${\displaystyle {x_1}^3+{x_2}^3=}$${\displaystyle (x_1+x_2)({x_1}^2-x_1{x}_2+{x_2}^2)=}$${\displaystyle \sqrt{2}\cdot (10-(-4))=}$${\displaystyle \sqrt{2}\cdot 14=}$${\displaystyle 14\sqrt{2}}$.