Упр.1.2 ГДЗ Мерзляк 11 класс Углубленный уровень (Алгебра)

1) 5^(v3-1)^2 : (1/5)^(2v3);
2) ((v2)^(v6))^(v6);
3) ((10^(1/5))^(v5))^(-2v5).

1. $$5^{(\sqrt{3}-1)^2} : \left(\frac{1}{5}\right)^{2\sqrt{3}}$$
   
   $$\left(\frac{1}{5}\right)^{2\sqrt{3}}=5^{-2\sqrt{3}}$$
   
   $$5^{(\sqrt{3}-1)^2} : 5^{-2\sqrt{3}}=5^{(\sqrt{3}-1)^2+2\sqrt{3}}$$
   
   $$ (\sqrt{3}-1)^2+2\sqrt{3}=3-2\sqrt{3}+1+2\sqrt{3}=4$$
   
   $$5^4=625$$
2. $$(\sqrt{2})^{\sqrt{6}}\;^{\sqrt{6}}=(\sqrt{2})^{\sqrt{6}\cdot\sqrt{6}}=(\sqrt{2})^6$$
   
   $$\sqrt{6}\cdot\sqrt{6}=6$$
   
   $$(\sqrt{2})^6=(2^{1/2})^6=2^3=8$$
3. $$\left(\left(\sqrt[5]{10}\right)^{\sqrt{5}}\right)^{-2\sqrt{5}}$$
   
   $$\left(\sqrt[5]{10}\right)^{\sqrt{5}}=10^{\frac{1}{5}\sqrt{5}}$$
   
   $$\left(10^{\frac{\sqrt{5}}{5}}\right)^{-2\sqrt{5}}=10^{-\frac{2\sqrt{5}\cdot\sqrt{5}}{5}}=10^{-2}$$
   
   $$10^{-2}=0{,}01$$

Ответ

1) $$625$$; 2) $$8$$; 3) $$0{,}01$$.