3. Найдите сумму, разность и произведение многочленов: – 3/20 y2 + 1/3 z3; 4/5 y2 – 5/12 z3.

Даны многочлены: $$-\frac{3}{20}y^2 + \frac{1}{3}z^3$$ и $$\frac{4}{5}y^2 - \frac{5}{12}z^3$$

1. Сумма многочленов:

$$\left(-\frac{3}{20}y^2 + \frac{1}{3}z^3\right) + \left(\frac{4}{5}y^2 - \frac{5}{12}z^3\right) = \left(-\frac{3}{20} + \frac{4}{5}\right)y^2 + \left(\frac{1}{3} - \frac{5}{12}\right)z^3$$

Приведем дроби к общему знаменателю:

$$= \left(-\frac{3}{20} + \frac{16}{20}\right)y^2 + \left(\frac{4}{12} - \frac{5}{12}\right)z^3 = \frac{13}{20}y^2 - \frac{1}{12}z^3$$

2. Разность многочленов:

$$\left(-\frac{3}{20}y^2 + \frac{1}{3}z^3\right) - \left(\frac{4}{5}y^2 - \frac{5}{12}z^3\right) = \left(-\frac{3}{20} - \frac{4}{5}\right)y^2 + \left(\frac{1}{3} + \frac{5}{12}\right)z^3$$ $$= \left(-\frac{3}{20} - \frac{16}{20}\right)y^2 + \left(\frac{4}{12} + \frac{5}{12}\right)z^3 = -\frac{19}{20}y^2 + \frac{9}{12}z^3 = -\frac{19}{20}y^2 + \frac{3}{4}z^3$$

3. Произведение многочленов:

$$\left(-\frac{3}{20}y^2 + \frac{1}{3}z^3\right) \cdot \left(\frac{4}{5}y^2 - \frac{5}{12}z^3\right) = -\frac{3}{20}y^2 \cdot \frac{4}{5}y^2 - \frac{3}{20}y^2 \cdot \left(-\frac{5}{12}z^3\right) + \frac{1}{3}z^3 \cdot \frac{4}{5}y^2 + \frac{1}{3}z^3 \cdot \left(-\frac{5}{12}z^3\right)$$ $$= -\frac{12}{100}y^4 + \frac{15}{240}y^2z^3 + \frac{4}{15}y^2z^3 - \frac{5}{36}z^6 = -\frac{3}{25}y^4 + \frac{1}{16}y^2z^3 + \frac{4}{15}y^2z^3 - \frac{5}{36}z^6$$ $$= -\frac{3}{25}y^4 + \left(\frac{1}{16} + \frac{4}{15}\right)y^2z^3 - \frac{5}{36}z^6 = -\frac{3}{25}y^4 + \left(\frac{15}{240} + \frac{64}{240}\right)y^2z^3 - \frac{5}{36}z^6$$ $$= -\frac{3}{25}y^4 + \frac{79}{240}y^2z^3 - \frac{5}{36}z^6$$

Ответ:

• Сумма: $$\frac{13}{20}y^2 - \frac{1}{12}z^3$$
• Разность: $$-\frac{19}{20}y^2 + \frac{3}{4}z^3$$
• Произведение: $$-\frac{3}{25}y^4 + \frac{79}{240}y^2z^3 - \frac{5}{36}z^6$$