Найдите одночлен, равный произведению одночленов:
а) $1\frac{1}{5}a^2b^3 * 1\frac{1}{9}ab^2$;
б) $(-1\frac{2}{3})b^2c^3 * (-\frac{2}{15})b^2c^2$;
в) $\frac{1}{2}ck^2 * \frac{2}{3}ck$;
г) $1\frac{2}{3}k^3p^2 * (-1\frac{1}{5})kp^2$;
д) $(-2\frac{1}{4})p^2x^2 * 1\frac{1}{3}px^3$;
е) $(-\frac{9}{11})x^2y^3 * (-1\frac{2}{9})xy$;
ж) $(-1\frac{2}{3})a^2x^3 * (-\frac{3}{5})a^2x^4$;
з) $(-2\frac{5}{6})a^3c^2 * 1\frac{2}{3}ac^2$.
$1\frac{1}{5}a^2b^3 * 1\frac{1}{9}ab^2 = (\frac{6}{5} * \frac{10}{9})a^{2 + 1}b^{3 + 2} = (\frac{2}{1} * \frac{2}{3})a^{3}b^{5} = \frac{4}{3}a^{3}b^{5} = 1\frac{1}{3}a^{3}b^{5}$
$(-1\frac{2}{3})b^2c^3 * (-\frac{2}{15})b^2c^2 = (\frac{5}{3} * \frac{2}{15})b^{2 + 2}c^{3 + 2} = (\frac{1}{3} * \frac{2}{3})b^{4}c^{5} = \frac{2}{9}b^{4}c^{5}$
$\frac{1}{2}ck^2 * \frac{2}{3}ck = (\frac{1}{2} * \frac{2}{3})c^{1 + 1}k^{2 + 1} = \frac{1}{3}c^{2}k^{3}$
$1\frac{2}{3}k^3p^2 * (-1\frac{1}{5})kp^2 = -(\frac{5}{3} * \frac{6}{5})k^{3 + 1}p^{2 + 2} = -(\frac{1}{1} * \frac{2}{1})k^{4}p^{4} = -2k^{4}p^{4}$
$(-2\frac{1}{4})p^2x^2 * 1\frac{1}{3}px^3 = -(\frac{9}{4} * \frac{4}{3})p^{2 + 1}x^{2 + 3} = -(\frac{3}{1} * \frac{1}{1})p^{3}x^{5} = -3p^3x^5$
$(-\frac{9}{11})x^2y^3 * (-1\frac{2}{9})xy = (\frac{9}{11} * \frac{11}{9})x^{2 + 1}y^{3 + 1} = x^3y^4$
$(-1\frac{2}{3})a^2x^3 * (-\frac{3}{5})a^2x^4 = (\frac{5}{3} * \frac{3}{5})a^{4}x^{7} = a^4x^7$
$(-2\frac{5}{6})a^3c^2 * 1\frac{2}{3}ac^2 = -(\frac{17}{6} * \frac{5}{3})a^{3 + 1}c^{2 + 2} = -\frac{85}{18}a^{4}c^{4} = -4\frac{13}{18}a^{4}c^{4}$
Пожауйста, оцените решение
