Упр.23.16 ГДЗ Мерзляк Поляков 9 класс (Алгебра)
1) a-b; 2) a^1,5-b^1,5; 3) m^0,6-8n^1,8; 4) x^(6/7)-6.
Используем формулу разности кубов:
$$u^3-v^3=(u-v)(u^2+uv+v^2).$$
$$a-b=a^{\frac13\cdot 3}-b^{\frac13\cdot 3}=(a^{\frac13}-b^{\frac13})(a^{\frac23}+a^{\frac13}b^{\frac13}+b^{\frac23})$$
$$=(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}).$$
$$a^{1,5}-b^{1,5}=a^{\frac32}-b^{\frac32}=(a^{\frac12}-b^{\frac12})(a+b^{\frac12}a^{\frac12}+b)$$
$$=(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b).$$
$$m^{0,6}-8n^{1,8}=m^{\frac35}-2^3n^{\frac95}=(m^{\frac15}-2n^{\frac35})(m^{\frac25}+2m^{\frac15}n^{\frac35}+4n^{\frac65})$$
$$=(\sqrt[5]{m}-2\sqrt[5]{n^3})(\sqrt[5]{m^2}+2\sqrt[5]{mn^3}+4\sqrt[5]{n^6}).$$
$$x^{\frac67}-6=x^{\frac67}-6^{\frac13\cdot 3}=(x^{\frac27}-\sqrt[3]{6})(x^{\frac47}+x^{\frac27}\sqrt[3]{6}+\sqrt[3]{36})$$
$$=(\sqrt[7]{x^2}-\sqrt[3]{6})(\sqrt[7]{x^4}+\sqrt[3]{6}\sqrt[7]{x^2}+\sqrt[3]{36}).$$
Ответ
1) $$(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2})$$
2) $$(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)$$
3) $$(\sqrt[5]{m}-2\sqrt[5]{n^3})(\sqrt[5]{m^2}+2\sqrt[5]{mn^3}+4\sqrt[5]{n^6})$$
4) $$(\sqrt[7]{x^2}-\sqrt[3]{6})(\sqrt[7]{x^4}+\sqrt[3]{6}\sqrt[7]{x^2}+\sqrt[3]{36})$$
