Question

[2,(5)] la puterea a 4-a · [ 2,( 5)] la puterea a 5-a 

(49/9) la puterea 21 : {[5,(4)] la puterea a 4-a} la puterea a 5-a

(75/19) la puterea a 3-a · ( 38/25) la puterea a 3-a

(0,12) la puterea a 7-a · (8intregi 1/3) la puterea a 7-a.

Answer 1

1)
$[2,(5)]^{4} * [2,(5)]^{5} = [2,(5)]^{4+5}= [2,(5)]^{9} = (\frac{25-2}{9} )^{9} = (\frac{23}{9}) ^{9}$

2)
$[( \frac{49}{9})^{21}: (\frac{54-5}{9}) ^{4}]^{5} = [( \frac{49}{9})^{21}:(\frac{49}{9}) ^{4}]^{5} =$

$=[(\frac{49}{9}) ^{21-4}]^{5} = [(\frac{49}{9}) ^{17}]^{5} = (\frac{49}{9}) ^{17*5} =(\frac{49}{9}) ^{85}$

3)
$(\frac{75}{19})^{3}*(\frac{38}{25})^{3} = (\frac{75}{19}* \frac{38}{25})^{3}=(\frac{75*38}{19*25})^{3} =(\frac{3*2}{1*1})^{3}= 6^{3}=216$

4)
$(0,12)^{7}*(8 \frac{1}{3})^{7}= (\frac{12}{100})^{7}*(\frac{25}{3})^{7}=(\frac{12}{100}*\frac{25}{3})^{7} =(\frac{4*1}{4*1})^{7}=(\frac{1}{1})^{7} = 1^{7}=1$