Question

VA ROG AM NEVOIE EXTREM DE URGENT P.S DE LA G INCOLO ​

Answer 1

Răspuns:

Explicație pas cu pas:

g. $\frac{2^4^9}{8^1^7} = \frac{2^4^9}{(2^3)^1^7=\frac{2^4^9}{2^5^1} } =\frac{1}{2^2} =2^-^2$

h. $\frac{8^2^5}{16^2^0} =\frac{(2^3)^2^5}{(2^4)^2^0} =\frac{2^7^5}{2^8^0} =\frac{1}{2^5} =2^-^5$

i. $\frac{4^2^6}{32^1^1} =\frac{(2^2)^2^6}{(2^5)^1^1} =\frac{2^5^2}{2^5^5} =\frac{1}{2^3}=2^-^3$

j. $\frac{9^1^7}{3^3^7} =\frac{(3^2)^1^7}{3^3^7} =\frac{3^3^4}{3^3^7} =\frac{1}{3^3} = 3^-^3$

k.$\frac{27^3^0}{3^9^1} =\frac{(3^3)^3^0}{3^9^1} =\frac{3^9^0}{3^9^1} =\frac{1}{3^1} =3^-^1$

l. $\frac{81^1^9}{9^3^9} =\frac{(9^2)^1^9}{9^3^9} =\frac{9^3^8}{9^3^9} =\frac{1}{9^1} =9^-^1$