Упр.39.42 ГДЗ Мордкович 7 класс (Алгебра)

а) $$\frac{a^6}{8}-b^9$$; б) $$\frac{8a^3}{27}+\frac{x^9}{64}$$; в) $$\frac{x^3}{125}+y^6$$; г) $$\frac{64m^3}{729}-\frac{343n^6}{1000}$$.

1. $$\frac18a^6-b^9=\left(\frac12a^2\right)^3-(b^3)^3$$
   По формуле разности кубов $$x^3-y^3=(x-y)(x^2+xy+y^2)$$ получаем:
   $$\frac18a^6-b^9=\left(\frac12a^2-b^3\right)\left(\frac14a^4+\frac12a^2b^3+b^6\right).$$
2. $$\frac{8}{27}a^3+\frac{1}{64}x^9=\left(\frac23a\right)^3+\left(\frac14x^3\right)^3$$
   По формуле суммы кубов $$x^3+y^3=(x+y)(x^2-xy+y^2)$$:
   $$\frac{8}{27}a^3+\frac{1}{64}x^9=\left(\frac23a+\frac14x^3\right)\left(\frac49a^2-\frac16ax^3+\frac1{16}x^6\right).$$
3. $$\frac{1}{125}x^3+y^6=\left(\frac15x\right)^3+(y^2)^3$$
   Тогда
   $$\frac{1}{125}x^3+y^6=\left(\frac15x+y^2\right)\left(\frac1{25}x^2-\frac15xy^2+y^4\right).$$
4. $$\frac{64}{729}m^3-\frac{343}{1000}n^6=\left(\frac49m\right)^3-\left(\frac7{10}n^2\right)^3$$
   По формуле разности кубов:
   $$\frac{64}{729}m^3-\frac{343}{1000}n^6=\left(\frac49m-\frac7{10}n^2\right)\left(\frac{16}{81}m^2+\frac{14}{45}mn^2+\frac{49}{100}n^4\right).$$

Ответ

а) $$\left(\frac12a^2-b^3\right)\left(\frac14a^4+\frac12a^2b^3+b^6\right)$$

б) $$\left(\frac23a+\frac14x^3\right)\left(\frac49a^2-\frac16ax^3+\frac1{16}x^6\right)$$

в) $$\left(\frac15x+y^2\right)\left(\frac1{25}x^2-\frac15xy^2+y^4\right)$$

г) $$\left(\frac49m-\frac7{10}n^2\right)\left(\frac{16}{81}m^2+\frac{14}{45}mn^2+\frac{49}{100}n^4\right)$$