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

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

a) \(< \)

b) \(>\)

c) \(< \)

d) \(>\)

e) \(< \)

g) \(>\)

h) \(>\)

k) \(>\)

4/5 > 3/7

11/15 < 12/6

-3/7 < -4/9

-5/8 < 4/-7

a, Ta có: 3³⁹ < 3⁴⁰ = (3²)²⁰ = 9²⁰

11²¹ > 11²⁰

Vì 9 < 11 => 9²⁰ < 11²⁰ => 3³⁹ < 11²¹

b, Ta có: 12⁴⁵ - 12⁴⁴ = 12⁴⁴(12 - 1) = 12⁴⁴.11

12⁴⁴ - 12⁴³ = 12⁴³(12 - 1) = 12⁴³.11

Vì 12⁴⁴ > 12⁴³ => 12⁴⁴.11 > 12⁴³.11 => 12⁴⁵ - 12⁴⁴ > 12⁴⁴ - 12⁴³

c, Đặt A = 1+2+2²+....+2⁹

2A = 2+2²+2³+...+2¹⁰

2A-A=2¹⁰-1

A=2¹⁰-1

Ta có: 5.2⁸ = 2⁸.4 > 2⁸.4 - 1 = 2¹⁰ - 1

Vậy A < 5.2⁸

`@` `\text {Ans}`

`\downarrow`

`1,`

`a)`

`3^12` và `5^8`

\(3^{12}=\left(3^3\right)^4=9^4\)

\(5^8=\left(5^2\right)^4=25^4\)

Vì `9 < 25` `=> 25^4 > 9^4`

`=> 3^12 > 5^8`

Vậy, `3^12 > 5^8`

`b)`

`(0,6)^9` và `(-0,9)^6`

\(\left(0,6\right)^9=\left(0,6^3\right)^3=\left(0,216\right)^3\)

\(\left(-0,9\right)^6=\left[\left(-0,9\right)^2\right]^3=\left(0,81\right)^3\)

Vì `0,81 > 0,216 => (0,81)^3 > (0,216)^3`

`=> (0,6)^9 < (-0,9)^6`

Vậy, `(0,6)^9<(-0,9)^6`

1.a) Có 3¹² = 3^3.4 = 27⁴ ;

5⁸ = 5^2.4 = 25⁴ 

Dễ thấy 27⁴ > 25⁴ nên 3¹² > 5⁸

b) Có  \(0,6^9=\dfrac{6^9}{10^9}=\dfrac{6^{3.3}}{10^9}=\dfrac{216^3}{10^9}\) 

mà \(\left(-0,9\right)^6=0,9^6=\dfrac{9^6}{10^6}=\dfrac{9^6.10^3}{10^9}=\dfrac{9^{2.3}.10^3}{10^9}=\dfrac{81^3.10^3}{10^9}=\dfrac{810^3}{10^9}\)

Dễ thấy \(\dfrac{216^3}{10^9}< \dfrac{810^3}{10^9}\Rightarrow0,6^9< \left(-0,9\right)^6\)

`2,`

`a)`

`31^5` và `17^7`

`31^5 < 32^5 = (2^5)^5 = 2^25`

`17^7 > 16^7 = (2^4)^7 = 2^28`

Vì `28 > 25 => 2^28 > 2^25`

`=> 31^5 < 17^7`

Vậy, `31^5 < 17^7`

`b)`

`8^12` và `12^8`

`8^12 = (8^3)^4 = 512^4`

`12^8 = (12^2)^4 = 144^4`

Vì `512 > 144 => 512^4 > 144^4`

`=> 12^8 < 8^12`

Vậy, `12^8 < 8^12.`

\(a,\dfrac{5}{3}>\dfrac{3}{5};b,\dfrac{6}{11}< \dfrac{9}{5};c,\dfrac{6}{11}=\dfrac{6}{11};d,\dfrac{8}{9}< \dfrac{8}{5}\)

 \(a,>\)

\(b,< \)

\(c,=\)

\(d,< \)

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