Упр.71 ГДЗ Алимов 10-11 класс (Алгебра)

1. $$\frac{10^{2+\sqrt{7}}}{2^{2+\sqrt{7}}\cdot 5^{1+\sqrt{7}}}$$
2. $$\frac{6^{3+\sqrt{5}}}{2^{2+\sqrt{5}}+3^{1+\sqrt{5}}}$$
3. $$(25^{1+\sqrt{2}}-5^{2\sqrt{2}})\cdot 5^{-1-2\sqrt{2}}$$
4. $$(2^{2\sqrt{3}}-4^{\sqrt{3}-1})\cdot 2^{-2\sqrt{3}}$$

1. $$\frac{10^{2+\sqrt7}}{2^{2+\sqrt7}\cdot 5^{1+\sqrt7}}=\frac{(2\cdot 5)^{2+\sqrt7}}{2^{2+\sqrt7}\cdot 5^{1+\sqrt7}}=\frac{2^{2+\sqrt7}\cdot 5^{2+\sqrt7}}{2^{2+\sqrt7}\cdot 5^{1+\sqrt7}}=5^{(2+\sqrt7)-(1+\sqrt7)}=5.$$

2. $$\frac{6^{3+\sqrt5}}{2^{2+\sqrt5}\cdot 3^{1+\sqrt5}}=\frac{(2\cdot 3)^{3+\sqrt5}}{2^{2+\sqrt5}\cdot 3^{1+\sqrt5}}=\frac{2^{3+\sqrt5}\cdot 3^{3+\sqrt5}}{2^{2+\sqrt5}\cdot 3^{1+\sqrt5}}=2^{(3+\sqrt5)-(2+\sqrt5)}\cdot 3^{(3+\sqrt5)-(1+\sqrt5)}=2\cdot 9=18.$$

3. $$\left(25^{1+\sqrt2}-5^{2\sqrt2}\right)\cdot 5^{-1-2\sqrt2}=\left((5^2)^{1+\sqrt2}-5^{2\sqrt2}\right)\cdot 5^{-1-2\sqrt2}$$
   $$=\left(5^{2+2\sqrt2}-5^{2\sqrt2}\right)\cdot 5^{-1-2\sqrt2}=5^{2+2\sqrt2-1-2\sqrt2}-5^{2\sqrt2-1-2\sqrt2}=5- \frac15=\frac45.$$

4. $$\left(2^{2\sqrt3}-4^{\sqrt3-1}\right)\cdot 2^{-2\sqrt3}=\left(2^{2\sqrt3}-(2^2)^{\sqrt3-1}\right)\cdot 2^{-2\sqrt3}$$
   $$=\left(2^{2\sqrt3}-2^{2\sqrt3-2}\right)\cdot 2^{-2\sqrt3}=2^{2\sqrt3-2\sqrt3}-2^{2\sqrt3-2-2\sqrt3}=1-\frac14=\frac34.$$

Ответ: 1) $$5$$; 2) $$18$$; 3) $$\frac45$$; 4) $$\frac34$$.