Упр.399 ГДЗ Никольский Потапов 9 класс (Алгебра)
- а) $$(x^{1/3}-y^{1/2})^3$$; б) $$(m^{1/2}+n^{2/3})^3$$; в) $$(a^{1/2}-b^{1/3})^3$$.
Используем формулу куба суммы и куба разности:
$$(u-v)^3=u^3-3u^2v+3uv^2-v^3,$$
$$(u+v)^3=u^3+3u^2v+3uv^2+v^3.$$
$$\left(x^{\frac13}-y^{\frac12}\right)^3 =\left(x^{\frac13}\right)^3-3\left(x^{\frac13}\right)^2y^{\frac12}+3x^{\frac13}\left(y^{\frac12}\right)^2-\left(y^{\frac12}\right)^3$$
$$=x-3x^{\frac23}y^{\frac12}+3x^{\frac13}y-y^{\frac32}.$$
$$\left(m^{\frac12}+n^{\frac23}\right)^3 =\left(m^{\frac12}\right)^3+3\left(m^{\frac12}\right)^2n^{\frac23}+3m^{\frac12}\left(n^{\frac23}\right)^2+\left(n^{\frac23}\right)^3$$
$$=m^{\frac32}+3mn^{\frac23}+3m^{\frac12}n^{\frac43}+n^2.$$
$$\left(a^{\frac12}-b^{\frac13}\right)^3 =\left(a^{\frac12}\right)^3-3\left(a^{\frac12}\right)^2b^{\frac13}+3a^{\frac12}\left(b^{\frac13}\right)^2-\left(b^{\frac13}\right)^3$$
$$=a^{\frac32}-3ab^{\frac13}+3a^{\frac12}b^{\frac23}-b.$$
Ответ
а) $$x-3x^{\frac23}y^{\frac12}+3x^{\frac13}y-y^{\frac32}$$; б) $$m^{\frac32}+3mn^{\frac23}+3m^{\frac12}n^{\frac43}+n^2$$; в) $$a^{\frac32}-3ab^{\frac13}+3a^{\frac12}b^{\frac23}-b$$.
