a,

\(3^{34}=\left(3^{17}\right)^2=129140163^2\)

\(5^{20}=\left(5^{10}\right)^2=9765625^2\)

Vậy..........

Đùa chút thui

c,\(3^{23}=3^{21}.3^2=\left(3^3\right)^7.9=27^7.9\)

\(5^{15}=\)\(5^{14}.5=\left(5^2\right)^7.5=25^7.5\)

\(27^7>25^7\)và \(9>5\)

nên \(3^{23}>5^{15}\)

b,

5^299 < 5^300 = (5^2)^150 = 25^150

3^501 = (3^3)^167 = 27^167

=> 27^167 > 25^150 => 3^501 > 5^299

Bye bye

Chúc bạn học tốt

Theo dõi mik nha<3

Ta có: 3³⁴>3³⁰=27¹⁰

5²⁰=25¹⁰

Vì 27¹⁰>25¹⁰ nên 3³⁴>5²⁰

a, 9^5>27^3

​b,3^200>2^300

​c, 32^11<17^14

a) \(\dfrac{-1}{20}=\dfrac{-7}{140}\)

\(\dfrac{5}{7}=\dfrac{100}{140}\)

mà -7<100

nên \(-\dfrac{1}{20}< \dfrac{5}{7}\)

b) \(\dfrac{216}{217}< 1\)

\(1< \dfrac{1164}{1163}\)

nên \(\dfrac{216}{217}< \dfrac{1164}{1163}\)

c) \(\dfrac{-12}{17}=\dfrac{-180}{255}\)

\(\dfrac{-14}{15}=\dfrac{-238}{255}\)

mà -180>-238

nên \(-\dfrac{12}{17}>\dfrac{-14}{15}\)

d) \(\dfrac{27}{29}>0\)

\(0>-\dfrac{2727}{2929}\)

nên \(\dfrac{27}{29}>-\dfrac{2727}{2929}\)

ta co

5^2=25

2^5= 32

32>25

suy ra 5^2<2^5

ta có

3^25 =3^23.3^2

ma 3^23<3^23.3^2

suy ra 3^25>3^23

ta co

3^37 =3^35.3^2        

        = 3^35.9

vì 3^35.9<10.3^35

suy ra 3^37<10.3^35