B Вычислить √(27.3-4)2 √16.(2-3)2 (108)2.100-6 (10-10.1006)-4 √6-4.6-9 √7-2.7-8 7-11 6-4 3-4.2-6 3-2.5-3 3-10.98 (-3)2

Решения:

1. $$√(27 \cdot 3^{-4})^2 = (27 \cdot 3^{-4}) = (3^3 \cdot 3^{-4}) = 3^{3-4} = 3^{-1} = {1\over 3}$$


2. $$√16 \cdot (2^{-3})^2 = √16 \cdot 2^{-6} = 4 \cdot {1\over 2^6} = 4 \cdot {1\over 64} = {4\over 64} = {1\over 16}$$


3. $$(10^8)^2 \cdot 100^{-6} = 10^{16} \cdot 10^{-12} = 10^{16-12} = 10^4 = 10000$$


4. $$(10^{-10} \cdot 100^6)^{-4} = (10^{-10} \cdot 10^{12})^{-4} = (10^{-10+12})^{-4} = (10^2)^{-4} = 10^{-8} = {1\over 10^8} = {1\over 100000000}$$


5. $$√6^{-4} \cdot 6^{-9} = √6^{-4-9} = √6^{-13} = 6^{-{13\over 2}} = {1\over 6^{{13\over 2}}}$$


6. $$√7^{-2} \cdot 7^{-8} = √7^{-2-8} = √7^{-10} = 7^{-5} = {1\over 7^5} = {1\over 16807}$$


7. $$7^{-11} = {1\over 7^{11}} = {1\over 1977326743}$$


8. $$6^{-4} = {1\over 6^4} = {1\over 1296}$$


9. $$3^{-4} \cdot 2^{-6} = {1\over 3^4} \cdot {1\over 2^6} = {1\over 81} \cdot {1\over 64} = {1\over 5184}$$


10. $$3^{-2} \cdot 5^{-3} = {1\over 3^2} \cdot {1\over 5^3} = {1\over 9} \cdot {1\over 125} = {1\over 1125}$$


11. $$3^{-10} \cdot 98 = {1\over 3^{10}} \cdot 98 = {98\over 59049}$$


12. $$(-3)^2 = 9$$

Ответ: 1) $${1\over 3}$$, 2) $${1\over 16}$$, 3) $$10000$$, 4) $${1\over 100000000}$$, 5) $${1\over 6^{{13\over 2}}}$$, 6) $${1\over 16807}$$, 7) $${1\over 1977326743}$$, 8) $${1\over 1296}$$, 9) $${1\over 5184}$$, 10) $${1\over 1125}$$, 11) $${98\over 59049}$$, 12) $$9$$