Задание 216

Решение:

• $${\largeа)}\ 1\frac{1}{5}a^2b^3\cdot1\frac{1}{9}ab^2=\left(\frac{6}{5}\cdot\frac{10}{9}\right)a^{2\ +\ 1}b^{3\ +\ 2}=\frac{4}{3}a^3b^5=1\frac{1}{3}a^3b^5$$

• $${\largeб)}\ \left(-1\frac{2}{3}\right)b^2c^3\cdot\left(-\frac{2}{15}\right)b^2c^2=\left(-\frac{5}{3}\cdot\left(-\frac{2}{15}\right)\right)b^{2\ +\ 2}c^{3\ +\ 2}=\frac{2}{9}b^4c^5$$

• $${\largeв)}\ \frac{1}{2}ck^2\cdot\frac{2}{3}ck=\left(\frac{1}{2}\cdot\frac{2}{3}\right)c^{1\ +\ 1}k^{2\ +\ 1}=\frac{1}{3}c^2k^3$$

• $${\largeг)}\ 1\frac{2}{3}k^3p^2\cdot\left(-1\frac{1}{5}\right)kp^2=\left(\frac{5}{3}\cdot\left(-\frac{6}{5}\right)\right)k^{3\ +\ 1}p^{2\ +\ 2}=(-2)k^4p^4$$

• $${\largeд)}\ \left(-2\frac{1}{4}\right)p^2x^2\cdot1\frac{1}{3}px^3=\left(-\frac{9}{4}\cdot\frac{4}{3}\right)p^{2\ +\ 1}x^{2\ +\ 3}=(-3)p^3x^5$$

• $${\largeе)}\ \left(-\frac{9}{11}\right)x^2y^3\cdot\left(-1\frac{2}{9}\right)xy=\left(-\frac{9}{11}\cdot\left(-\frac{11}{9}\right)\right)x^{2\ +\ 1}y^{3\ +\ 1}=x^3y^4$$

• $${\largeж)}\ \left(-1\frac{2}{3}\right)a^2x^3\cdot\left(-\frac{3}{5}\right)a^2x^4=\left(-\frac{5}{3}\cdot\left(-\frac{3}{5}\right)\right)a^{2\ +\ 2}x^{3\ +\ 4}=a^4x^7$$

• $${\largeз)}\ \left(-2\frac{5}{6}\right)a^3c^2\cdot1\frac{2}{3}ac^2=\left(-\frac{17}{6}\cdot\frac{5}{3}\right)a^{3\ +\ 1}c^{2\ +\ 2}=\left(-\frac{85}{18}\right)a^4c^4=\left(-4\frac{13}{18}\right)a^4c^4$$