Задача 55039 ...
Условие
Решение
[m]x_{1}+x_{2}+x_{3}=6[/m]
[m]x_{1}\cdot x_{2}+ x_{2}\cdot x_{3}+x_{1}\cdot x_{3}=12[/m]
[m]x_{1}\cdot x_{2}\cdot x_{3}=18[/m]
Пусть корни нового многочлена:
[m]t_{1}=\frac{1}{x_{1}}[/m]
[m]t_{2}=\frac{1}{x_{2}}[/m]
[m]t_{3}=\frac{1}{x_{3}}[/m]
Тогда
[m]t_{1}+t_{2}+t_{3}=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}=\frac{x_{1}\cdot x_{2}+ x_{2}\cdot x_{3}+x_{1}\cdot x_{3}}{x_{1}\cdot x_{2}\cdot x_{3}}=\frac{12}{18}=\frac{2}{3}[/m] ⇒ [m]-\frac{b}{a}=\frac{2}{3}[/m]
[m]t_{1}\cdot t_{2}+ t_{2}\cdot t_{3}+t_{1}\cdot t_{3}= \frac{1}{x_{1}}\cdot \frac{1}{x_{2}}+\frac{1}{x_{2}}\cdot\frac{1}{x_{3}}+\frac{1}{x_{1}}\cdot \frac{1}{x_{3}}=\frac{x_{3}+x_{1}+x_{2}}{x_{1}\cdot x_{2}\cdot x_{3}}=\frac{6}{18}=\frac{1}{3}[/m]⇒ [m]\frac{c}{a}=\frac{1}{3}[/m]
[m]t_{1}\cdot t_{2}\cdot t_{3}=\frac{1}{x_{1}}\cdot \frac{1}{x_{2}}\cdot \frac{1}{x_{3}}=\frac{1}{x_{1}\cdot x_{2}\cdot x_{3}}=\frac{1}{18}[/m]⇒ [m]-\frac{d}{a}=\frac{1}{18}[/m]
[m]x^3-\frac{2}{3}x^2+\frac{1}{3}x-\frac{1}{18}[/m]
Умножаем на 2:
[m]2x^3-\frac{4}{3}x^2+\frac{2}{3}x-\frac{1}{9}[/m] - о т в е т.