Question

Exercitiul 12!! Puteti trimite poze cu exercitiul efectuat pe un caiet sau pe o foaie sa va fie mai usor!
Urgent!!
Dau coroana chiar dupa rezolavare celui ce va rezolva corect

Answer 1

$12.a)\frac{6^{-10} }{81^{-2} \cdot 32^{-2} }=\\ \\ \frac{2^{-10}\cdot3^{-10}}{(3^{4})^{-2}\cdot (2^{5})^{-2}}= \\ \\ =\frac{2^{-10}\cdot 3^{-10}}{(3^{-8})\cdot (2^{-10})}=\\ \\= 3^{-10+8}\\ \\ =3^{-2}=\frac{1}{9}$

$b)(\frac{1}{5})^{-25}\cdot25^{-6}\cdot125^{-4} =\\ \\ \frac{1}{5^{-25} } \cdot5^{2\cdot-6} \cdot5^{3\cdot-4} =\\ \\ \frac{1}{5^{-25} } \cdot5^{-12}\cdot^{-12} =\\ \\ \frac{1}{5^{-25} }\cdot5^{-24}=\\ \\ 5^{25-24} \\ \\ =5$

$c)\\ \frac{20^{-4} \cdot15^{-3}}{30^{-7}}=\\ \\ \frac{(2^{2} \cdot5)^{-4}\cdot(3\cdot5)^{-3}}{(3\cdot5\cdot2)^{-7}}=\\ \\ \frac{2^{-8}\cdot5^{-4}\cdot3^{-3}\cdot5^{-3}}{3^{-7}\cdot5^{-7}\cdot2^{-7}}=\\ \\ \frac{2^{-8}\cdot3^{-3}\cdot5^{-7}}{3^{-7}\cdot5^{-7}\cdot2^{-7}} =\\ \\ =2^{-8}\cdot2^{7} \cdot3^{-3}\cdot3^{7}\cdot5^{-7}\cdot5^{7}=\\ \\= 2^{-1}\cdot3^{4}=\frac{81}{2}$

$f)\frac{(2^{3})^{5}\cdot(2^{-6)^{2}}}{4^{2} } =\\ \\ \frac{2^{15}\cdot2^{-12}}{(2^{2}) ^{2}} =\\ \\ \frac{2^{3}}{2^{4} }=2^{3-4}=2^{-1}=\frac{1}{2}$

Answer 2

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