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) 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: \(8242408>91809\)

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

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

\(a,\)\(\text{Ta có: }\) \(3^{200}=\left(3^2\right)^{100}=9^{100}\left(1\right)\)

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

\(\text{Từ (1) và (2) }\)\(\Rightarrow3^{200}>2^{300}\)

a, 3^200= (3^2)^100= 9^100

2^300= (2^3)^100= 8^100

Vì 9^100>8^100 nên 3^200>2^300

b, 125^5= (5^3)^5= 5^15

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

Vì 5^15>5^14 nên 125^5>25^7

a ) 2²⁰⁰<8³⁰⁰

b ) 25²⁰⁰>5³⁰⁰

c ) 4²¹<64⁷

a)\(8^{300}=\left(2^3\right)^{300}=2^{900}\)

Vì \(200< 900\Rightarrow2^{200}< 8^{300}\)

b)\(25^{200}=\left(5^2\right)^{200}=5^{400}\)

Vì \(400>300\Rightarrow25^{200}>5^{300}\)

c)\(64^7=\left(4^3\right)^7=4^{21}\)

Vì \(4^{21}=4^{21}\Rightarrow4^{21}=64^7\)

a) \(2^{200}=2^{200}\)

\(8^{300}=\left(2^3\right)^{300}=2^{900}\)

vi \(2^{200}< 2^{900}\)nen \(2^{200}< 8^{300}\)

b) \(25^{200}=\left(5^2\right)^{200}=5^{400}\)

\(5^{300}=5^{300}\)

vi \(5^{400}>5^{300}\)nen \(25^{200}>5^{300}\)

c) \(4^{21}=\left(2^2\right)^{21}=2^{42}\)

\(64^7=\left(2^6\right)^7=2^{42}\)

vi \(2^{42}=2^{42}\)nen \(4^{21}=64^7\)

Bài 1:

a: Sửa đề: 1/3^200

1/2^300=(1/8)^100

1/3^200=(1/9)^100

mà 1/8>1/9

nên 1/2^300>1/3^200

b: 1/5^199>1/5^200=1/25^100

1/3^300=1/27^100

mà 25^100<27^100

nên 1/5^199>1/3^300

a.

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

Vậy \(3^{200}>2^{300}\)

b.

\(5^{200}=\left(5^2\right)^{100}=25^{100}< 32^{100}=\left(2^5\right)^{100}=2^{500}\)

Vậy \(5^{200}< 2^{500}\)

Ta có : \(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

\(\Rightarrow9^{100}>8^{100}\)

\(\Rightarrow3^{200}>2^{300}\)

Ta có : \(5^{200}=5^{2.100}=\left(5^2\right)^{100}=25^{100}\)

\(2^{500}=2^{5.100}=\left(2^5\right)^{100}=32^{100}\)

\(\Rightarrow25^{100}< 32^{100}\)

\(\Rightarrow5^{200}< 2^{500}\)

Ta có: 300²⁰⁰=(3.100)²⁰⁰=3²⁰⁰.100²⁰⁰=3^2.100.10^2.100=(3²)¹⁰⁰.100².100¹⁰⁰=9¹⁰⁰.100².100¹⁰⁰

200³⁰⁰=(2.100)³⁰⁰=2³⁰⁰.100³⁰⁰=2^3.100.10^3.100=(2³)¹⁰⁰.100²⁺¹.100¹⁰⁰=8¹⁰⁰.100.100².100¹⁰⁰

Ta thấy:8².100=8².10²=80²<81=9²

=>8².100<9²

Mà 8⁹⁸<9⁹⁸

=>8².100.8⁹⁸<9².9⁹⁸

=>8¹⁰⁰.100<9¹⁰⁰

=>8¹⁰⁰.100.100².100¹⁰⁰<9¹⁰⁰.100².100¹⁰⁰

=>200³⁰⁰<300²⁰⁰

ta có :

2³⁰⁰=(2³)¹⁰⁰=8¹⁰⁰

3²⁰⁰=(3²)¹⁰⁰=9¹⁰⁰

vì 8¹⁰⁰<9¹⁰⁰ nên 2³⁰⁰<3²⁰⁰

^{tick cái bạn}

Ta có: 300²⁰⁰=(3.100)²⁰⁰=3²⁰⁰.100²⁰⁰=3^2.100.10^2.100=(3²)¹⁰⁰.100².100¹⁰⁰=9¹⁰⁰.100².100¹⁰⁰

200³⁰⁰=(2.100)³⁰⁰=2³⁰⁰.100³⁰⁰=2^3.100.10^3.100=(2³)¹⁰⁰.100²⁺¹.100¹⁰⁰=8¹⁰⁰.100.100².100¹⁰⁰

Ta thấy:8².100=8².10²=80²<81²=9⁴

=>8².100<9⁴

Mà 8⁹⁸=8⁹⁶.8²=8⁹⁶.64=8⁹⁶.1,044⁹⁶<8⁹⁶.1,125⁹⁶=(8.1,125)⁹⁶=9⁹⁶

=>8⁹⁸<9⁹⁶

=>8².100.8⁹⁸<9⁴.9⁹⁶

=>8¹⁰⁰.100<9¹⁰⁰

=>8¹⁰⁰.100.100².100¹⁰⁰<9¹⁰⁰.100².100¹⁰⁰

=>200³⁰⁰<300²⁰⁰

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

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

3*27^8=3*3^24=3^25=243^5

3: 3^300=27^100

2^200=4^100

mà 27>4

nên 3^300>2^200

4: 15^2=3^2*5^2

81^3*125^3=3^12*5^9

=>15^2<81^3*125^3

6: 125^5=5^15

25^7=5^14

mà 15>14

nên 125^5>25^7

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