632. Представьте в виде многочлена: a) $$\frac{2}{7}x(1,4x^2 - 3,5y)$$; б) $$-\frac{1}{3}c^2(1,2d^2 - 6c)$$; в) $$\frac{1}{2}ab(\frac{2}{3}a^2 - \frac{3}{4}ab + \frac{4}{5}b^2)$$; г) $$-\frac{2}{5}a^2y^5(5ay^2 - \frac{1}{2}a^2y - \frac{5}{6}a^3)$$.

a) $$\frac{2}{7}x(1,4x^2 - 3,5y) = \frac{2}{7}x \cdot 1,4x^2 - \frac{2}{7}x \cdot 3,5y = 0,4x^3 - xу$$ б) $$-\frac{1}{3}c^2(1,2d^2 - 6c) = -\frac{1}{3}c^2 \cdot 1,2d^2 + \frac{1}{3}c^2 \cdot 6c = -0,4c^2d^2 + 2c^3$$ в) $$\frac{1}{2}ab(\frac{2}{3}a^2 - \frac{3}{4}ab + \frac{4}{5}b^2) = \frac{1}{2}ab \cdot \frac{2}{3}a^2 - \frac{1}{2}ab \cdot \frac{3}{4}ab + \frac{1}{2}ab \cdot \frac{4}{5}b^2 = \frac{1}{3}a^3b - \frac{3}{8}a^2b^2 + \frac{2}{5}ab^3$$ г) $$-\frac{2}{5}a^2y^5(5ay^2 - \frac{1}{2}a^2y - \frac{5}{6}a^3) = -\frac{2}{5}a^2y^5 \cdot 5ay^2 + \frac{2}{5}a^2y^5 \cdot \frac{1}{2}a^2y + \frac{2}{5}a^2y^5 \cdot \frac{5}{6}a^3 = -2a^3y^7 + \frac{1}{5}a^4y^6 + \frac{1}{3}a^5y^5$$ Ответы: a) $$0,4x^3 - xy$$ б) $$-0,4c^2d^2 + 2c^3$$ в) $$\frac{1}{3}a^3b - \frac{3}{8}a^2b^2 + \frac{2}{5}ab^3$$ г) $$-2a^3y^7 + \frac{1}{5}a^4y^6 + \frac{1}{3}a^5y^5$$