Упр.23.14 ГДЗ Мерзляк Поляков 9 класс (Алгебра)

1) a^5-b^5; 2) x^(1/6)-y^(1/6); 3) 5-c; 4) 16x^0,3-25y^(2/9).

1. $$a^5-b^5=\left(\sqrt{a^5}\right)^2-\left(\sqrt{b^5}\right)^2$$
   $$=\left(\sqrt{a^5}-\sqrt{b^5}\right)\left(\sqrt{a^5}+\sqrt{b^5}\right).$$

2. $$x^{1/6}-y^{1/6}=\left(x^{1/12}\right)^2-\left(y^{1/12}\right)^2$$
   $$=\left(x^{1/12}-y^{1/12}\right)\left(x^{1/12}+y^{1/12}\right).$$

3. $$5-c=\left(\sqrt{5}\right)^2-\left(\sqrt{c}\right)^2$$
   $$=\left(\sqrt{5}-\sqrt{c}\right)\left(\sqrt{5}+\sqrt{c}\right).$$

4. $$16x^{0,3}-25y^{2/9}=16x^{3/10}-25y^{2/9}$$
   $$=4^2\cdot x^{3/10}-5^2\cdot y^{2/9}$$
   $$=\left(4x^{3/20}-5y^{1/9}\right)\left(4x^{3/20}+5y^{1/9}\right).$$

Ответ

1) $$\left(\sqrt{a^5}-\sqrt{b^5}\right)\left(\sqrt{a^5}+\sqrt{b^5}\right);$$
2) $$\left(x^{1/12}-y^{1/12}\right)\left(x^{1/12}+y^{1/12}\right);$$
3) $$\left(\sqrt{5}-\sqrt{c}\right)\left(\sqrt{5}+\sqrt{c}\right);$$
4) $$\left(4x^{3/20}-5y^{1/9}\right)\left(4x^{3/20}+5y^{1/9}\right).$$