Представьте в виде многочлена выражение:
1) $(0,02p^3 + 20p^2k^4)^2$;
2) $(1\frac{1}{6}mn - \frac{4}{21}m^2n^5)^2$;
3) $-15(\frac{1}{3}a - \frac{1}{5}b)^2$;
4) $7x(x^3 - 2x)^2$;
5) $(5y - 2)^2(2y + 1)$;
6) $(10p - k)^2(10p + k)^2$.
$(0,02p^3 + 20p^2k^4)^2 = 0,0004p^6 + 0,8p^5k^4 + 400p^4k^8$
$(1\frac{1}{6}mn - \frac{4}{21}m^2n^5)^2 = (\frac{7}{6}mn - \frac{4}{21}m^2n^5)^2 = \frac{49}{36}m^2n^2 - 2 * \frac{7}{6} * \frac{4}{21}m^3n^6 + \frac{16}{441}m^4n^{10} = 1\frac{13}{36}m^2n^2 - 2 * \frac{1}{3} * \frac{2}{3}m^3n^6 + \frac{16}{441}m^4n^{10} = 1\frac{13}{36}m^2n^2 - \frac{4}{9}m^3n^6 + \frac{16}{441}m^4n^{10}$
$-15(\frac{1}{3}a - \frac{1}{5}b)^2 = -15(\frac{1}{9}a^2 - \frac{2}{15}ab + \frac{1}{25}b^2) = -\frac{5}{3}a^2 + 2ab - \frac{3}{5}b^2 = -1\frac{2}{3}a^2 + 2ab - \frac{3}{5}b^2$
$7x(x^3 - 2x)^2 = 7x(x^6 - 4x^4 + 4x^2) = 7x^7 - 28x^5 + 28x^3$
$(5y - 2)^2(2y + 1) = (25y^2 - 20y + 4)(2y + 1) = 50y^3 - 40y^2 + 8y + 25y^2 - 20y + 4 = 50y^3 + (-40y^2 + 25y^2) + (8y - 20y) + 4 = 50y^3 - 15y^2 - 12y + 4$
$(10p - k)^2(10p + k)^2 = ((10p - k)(10p + k))^2 = (100p^2 - k^2)^2 = 10000p^4 - 200p^2k^2 + k^4$
Пожауйста, оцените решение
