9^27=3^81 > 81^13  =3^52

5^14 =25^7 < 27^7 

10^30>9^30=3^90 > 2^100 (chú ý 3^3>2^4)

2^300=8^100 < 3^200=9^100

8^5=2^15=2^6.2^9 < 2^6.3^6  (chú ý 2^3<3^2)

3^450=(3^3)^150=27^150 > 5^300=(5^2)^150=25^150

a/ \(9^{27}=\left(3^2\right)^{27}=3^{54}\) và \(81^{13}=\left(3^4\right)^{13}=3^{52}\Rightarrow3^{54}>3^{52}\Rightarrow9^{27}>81^{13}\)

b/ \(5^{14}=\left(5^2\right)^7=25^7< 27^7\)

d/ \(2^{300}=\left(2^3\right)^{100}=8^{100}\) và \(3^{200}=\left(3^2\right)^{100}=9^{100}\Rightarrow8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)

f/ \(3^{450}=\left(3^3\right)^{150}=27^{150}\) và \(5^{300}=\left(5^2\right)^{150}=25^{150}\Rightarrow27^{150}>25^{150}\Rightarrow3^{450}>5^{300}\)

c/ \(10^{30}=\left(10^3\right)^{10}=1000^{10}\) và \(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\Rightarrow1000^{10}< 1024^{10}\Rightarrow10^{30}< 2^{100}\)

a, Chia hết cho 3 thì nhóm 2 số thành 1 cặp ; chia hết cho 7 thì nhóm 3 số thành 1 cặp

b, Đề phải là A = 2009.2011

Có :A = 2009.(2010+1) = 2009.2010+2009

= 2009.2010+2010-1 = 2010.(2009+1)-1 = 2010^2-1

Vì 2010^2-1 < 2010^2 = B => A < B

c, A = (3^3)^150 = 27^150

B = (5^2)^150 = 25^150

Vì 27^150 > 25^150 => A > B

k mk nha

a) <

b) =

c) =

d) =

e) =

f) =

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!

a/

\(9^5=\left(3^2\right)^5=3^{10}>3^9=\left(3^3\right)^3=27^3\)

b/ \(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}=\left(2^3\right)^{100}=2^{300}\)

c/

\(3.4^7=3.\left(2^2\right)^7=3.2^{14}>2.2^{14}=2^{15}=\left(2^3\right)^5=8^5\)

a: 99^20=9801^10<9999^10

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

1: 243^5=(3^5)^5=3^25

3*27^8=3*(3^3)^8=3^25

=>243^5=3*27^8

6: 125^5=(5^3)^5=5^15

25^7=(5^2)^7=5^14

=>125^5>25^7(15>14)

5: 78^12-78^11=78^11(78-1)=78^11*77

78^11-78^10=78^10*77

mà 11>10

nên 78^12-78^11>78^11-78^10

a) Ta có:

\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)

Mà: \(8< 9\)

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) Ta có:

\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)

Mà: \(243< 343\)

\(\Rightarrow243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

c) Ta có: 

\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)

\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)

Mà: \(2< 3\)

\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)

\(\Rightarrow8^5< 3\cdot4^7\)

d) Ta có:

\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)

\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)

Mà: \(8242408>91809\)

\(\Rightarrow8242408^{101}>91809^{101}\)

\(\Rightarrow202^{303}>303^{202}\)