Упр.39.39 ГДЗ Мордкович 7 класс (Алгебра)

а) 1a3/8 — 8b3/27;
б) 64c3/343+ 729d3/1000;
в) 125×3/512 — 216y3/343;
г) 1m3/729 + 125n3/216.

1. Используем формулы разложения суммы и разности кубов:

   $$a^3-b^3=(a-b)(a^2+ab+b^2),$$

   $$a^3+b^3=(a+b)(a^2-ab+b^2).$$

2. а) $$\frac18a^3-\frac{8}{27}b^3=\left(\frac12a\right)^3-\left(\frac23b\right)^3$$

   $$=\left(\frac12a-\frac23b\right)\left(\frac14a^2+\frac13ab+\frac49b^2\right).$$

3. б) $$\frac{64}{343}c^3+\frac{729}{1000}d^3=\left(\frac47c\right)^3+\left(\frac9{10}d\right)^3$$

   $$=\left(\frac47c+\frac9{10}d\right)\left(\frac{16}{49}c^2-\frac{18}{35}cd+\frac{81}{100}d^2\right).$$

4. в) $$\frac{125}{512}x^3-\frac{216}{343}y^3=\left(\frac58x\right)^3-\left(\frac67y\right)^3$$

   $$=\left(\frac58x-\frac67y\right)\left(\frac{25}{64}x^2+\frac{15}{28}xy+\frac{36}{49}y^2\right).$$

5. г) $$\frac{1}{729}m^3+\frac{125}{216}n^3=\left(\frac19m\right)^3+\left(\frac56n\right)^3$$

   $$=\left(\frac19m+\frac56n\right)\left(\frac{1}{81}m^2-\frac{5}{54}mn+\frac{25}{36}n^2\right).$$

Ответ

а) $$\left(\frac12a-\frac23b\right)\left(\frac14a^2+\frac13ab+\frac49b^2\right)$$
б) $$\left(\frac47c+\frac9{10}d\right)\left(\frac{16}{49}c^2-\frac{18}{35}cd+\frac{81}{100}d^2\right)$$
в) $$\left(\frac58x-\frac67y\right)\left(\frac{25}{64}x^2+\frac{15}{28}xy+\frac{36}{49}y^2\right)$$
г) $$\left(\frac19m+\frac56n\right)\left(\frac{1}{81}m^2-\frac{5}{54}mn+\frac{25}{36}n^2\right)$$