64 Пользуясь тождеством а²-b²= (a+b) (a - b), разложить на множители: 1) a²-b²; 2) -1; 3) a³-b³; 4) x-y; 5) 4a²-b²; 6) 0,01m-n

1. $$a^{\frac{1}{2}} - b^{\frac{1}{2}} = (a^{\frac{1}{4}} + b^{\frac{1}{4}})(a^{\frac{1}{4}} - b^{\frac{1}{4}})$$
2. $$y^{\frac{2}{3}} - 1 = (y^{\frac{1}{3}} + 1)(y^{\frac{1}{3}} - 1)$$
3. $$a^{\frac{1}{3}} - b^{\frac{1}{3}} = (a^{\frac{1}{6}} + b^{\frac{1}{6}})(a^{\frac{1}{6}} - b^{\frac{1}{6}})$$
4. $$x - y = (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$$
5. $$4a^{\frac{1}{2}} - b^{\frac{1}{2}} = (2a^{\frac{1}{4}} + b^{\frac{1}{4}})(2a^{\frac{1}{4}} - b^{\frac{1}{4}})$$
6. $$0.01m^{\frac{1}{6}} - n^{\frac{1}{6}} = (0.1m^{\frac{1}{12}} + n^{\frac{1}{12}})(0.1m^{\frac{1}{12}} - n^{\frac{1}{12}})$$

Ответ:

1. $$(a^{\frac{1}{4}} + b^{\frac{1}{4}})(a^{\frac{1}{4}} - b^{\frac{1}{4}})$$
2. $$(y^{\frac{1}{3}} + 1)(y^{\frac{1}{3}} - 1)$$
3. $$(a^{\frac{1}{6}} + b^{\frac{1}{6}})(a^{\frac{1}{6}} - b^{\frac{1}{6}})$$
4. $$(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$$
5. $$(2a^{\frac{1}{4}} + b^{\frac{1}{4}})(2a^{\frac{1}{4}} - b^{\frac{1}{4}})$$
6. $$(0.1m^{\frac{1}{12}} + n^{\frac{1}{12}})(0.1m^{\frac{1}{12}} - n^{\frac{1}{12}})$$