`#3107.101107`

a)

`64^150` và `4^450`

Ta có:

`64^150 = (4^3)^150 = 4^(3*150) = 4^450`

Vì `450 = 450 => 4^450 = 4^450 => 64^150 = 4^450`

Vậy, `64^150 = 4^450`

b)

`81^64` và `27^100`

Ta có:

`81^64 = (3^4)^64 = 3^(4*64) = 3^256`

`27^100 = (3^3)^100 = 3^(3*100) = 3^300`

Vì `256 < 300 => 3^256 < 3^300 => 81^64 < 27^100`

Vậy, `81^64 < 27^100`

c)

`125^1000` và `25^3000`

Ta có:

`125^1000 = (5^3)^1000 = 5^(3*1000) = 5^3000`

Vì `5 < 25 => 5^3000 < 25^3000 => 125^1000 < 25^3000`

Vậy, `125^1000 < 25^3000`

d)

`4^30` và `3^40`

Ta có:

`4^30 = 4^(3*10) = (4^3)^10 = 64^10`

`3^40 = 3^(4*10) = (3^4)^10 = 81^10`

Vì `64 < 81 => 64^10 < 81^10 => 4^30 < 3^40`

Vậy, `4^30 < 3^40`

m)

`2^5000` và `5^2000`

Ta có:

`2^5000 = 2^(5*1000) = (2^5)^1000 = 32^1000`

`5^2000 = 5^(2*1000) = (5^2)^1000 = 25^1000`

Vì `32 > 25 => 32^1000 > 25^1000 => 2^5000 > 5^2000`

Vậy, `2^5000 > 5^2000`

h)

`6^450` và `3^750`

Ta có:

`6^450 = 6^(150*3) = (6^3)^150 = 216^150`

`3^750 = 3^(150*5) = (3^5)^150 = 243^150`

Vì `216 < 243 => 216^150 < 243^150 => 6^450 < 3^750`

Vậy, `6^450 < 3^750`

0)

`333^444` và `444^333`

Ta có:

`333^444 = 333^(4*111) = (333^4)^111 = (3^4 *111^4)^111 = 81^111 * 111^444`

`444^333 = 444^(3*111) = (444^3)^111 = (4^3 * 111^3)^111 = 64^111 * 111^333`

Vì `81 > 64;` `111^444 > 111^333`

`=> 81^111 * 111^444 > 64^111 * 111^333`

Vậy, `333^444 > 444^333.`

a) Ta có:

\(64^{150}=\left(2^6\right)^{150}=2^{900}\)

\(4^{450}=\left(2^2\right)^{450}=2^{900}\)

Mà: \(2^{900}=2^{900}\Rightarrow64^{150}=4^{450}\)

b) Ta có:

\(81^{64}=\left(3^4\right)^{64}=3^{256}\)

\(27^{100}=\left(3^3\right)^{100}=3^{300}\)

Mà: \(3^{300}>3^{256}\Rightarrow27^{100}>81^{64}\)

c) Ta có: 

\(125^{1000}=\left(5^3\right)^{1000}=5^{3000}\)

Mà: \(25^{3000}>5^{3000}\Rightarrow25^{3000}>125^{1000}\)

d) Ta có:

\(4^{30}=\left(4^3\right)^{10}=64^{10}\)

\(3^{40}=\left(3^4\right)^{10}=81^{10}\)

Mà: \(81^{10}>64^{10}\Rightarrow3^{40}>4^{30}\)

m) Ta có:

\(2^{5000}=\left(2^5\right)^{1000}=32^{1000}\)

\(5^{2000}=\left(5^2\right)^{1000}=25^{1000}\)

Mà: \(25^{1000}< 32^{1000}\Rightarrow2^{5000}>5^{2000}\)

h) Ta có:

\(6^{450}=\left(6^3\right)^{150}=216^{150}\)

\(3^{750}=\left(3^5\right)^{150}=243^{150}\)

Mà: \(243^{150}>216^{150}\Rightarrow3^{750}>6^{450}\)

.... 

a) 4⁴⁵⁰ = (4³)¹⁵⁰ = 64¹⁵⁰

b) 81⁶⁴ = (3⁴)⁶⁴ = 3²⁵⁶

27¹⁰⁰ = (3³)¹⁰⁰ = 3³⁰⁰

Do 256 < 300 nên 3²⁵⁶ < 3³⁰⁰

Vậy 81⁶⁴ < 27¹⁰⁰

c) 125¹⁰⁰⁰ = (5³)¹⁰⁰⁰ = 5³⁰⁰⁰

Do 5 < 25 nên 5³⁰⁰⁰ < 25³⁰⁰⁰

Vậy 125¹⁰⁰⁰ < 25³⁰⁰⁰

d) 4³⁰ = (4³)¹⁰ = 64¹⁰

3⁴⁰ = (3⁴)¹⁰ = 81¹⁰

Do 64 < 81 nên 64¹⁰ < 81¹⁰

Vậy 4³⁰ < 3⁴⁰

m) 2⁵⁰⁰⁰ = (2⁵)¹⁰⁰⁰ = 32¹⁰⁰⁰

5²⁰⁰⁰ = (5²)¹⁰⁰⁰ = 25¹⁰⁰⁰

Do 32 > 25 nên 32¹⁰⁰⁰ > 25¹⁰⁰⁰

Vậy 2⁵⁰⁰⁰ > 5²⁰⁰⁰

h) 6⁴⁵⁰ = (6³)¹⁵⁰ = 216¹⁵⁰

3⁷⁵⁰ = (3⁵)¹⁵⁰ = 243¹⁵⁰

Do 216 < 243 nên 216¹⁵⁰ < 243¹⁵⁰

Vậy 6⁴⁵⁰ < 3⁷⁵⁰

o) 333⁴⁴⁴ = (333⁴)¹¹¹ = [(3.111)⁴]¹¹¹ = (3⁴.111⁴)¹¹¹ = (81.111⁴)¹¹¹

444³³³ = (444³)¹¹¹ = [(4.111)³]¹¹¹

= (4³.111³)¹¹¹ = (64.111³)¹¹¹

Do 81 > 64 ⇒ 81.111⁴ > 64.111⁴ (1)

Do 4 > 3 ⇒ 64.111⁴ > 64.111³ (2)

Từ (1) và (2) ⇒ 81.111⁴ > 64.111³

⇒ (81.111⁴)¹¹¹ > (64.111³)¹¹¹

Vậy 333⁴⁴⁴ > 444³³³

\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)

Vậy \(A< B\)

a) Ta có \(0,625^{200}=\left(\dfrac{5}{8}\right)^{200}\) và \(0,5^{1000}=\left(\dfrac{1}{2}\right)^{1000}=\left(\dfrac{1}{2}\right)^{5.200}\) \(=\left[\left(\dfrac{1}{2}\right)^5\right]^{200}\) \(=\left(\dfrac{1}{32}\right)^{200}\). Mà hiển nhiên \(\left(\dfrac{5}{8}\right)^{200}>\left(\dfrac{1}{32}\right)^{200}\) nên suy ra \(0,625^{200}>0,5^{1000}\)

b) Ta thấy \(\left(-32\right)^{27}< 0\) trong khi \(\left(-27\right)^{32}>0\) nên đương nhiên \(\left(-32\right)^{27}< \left(-27\right)^{32}\)

c) Ta thấy \(-\dfrac{3}{2}>-2\) nên \(\left(-\dfrac{3}{2}\right)^5>\left(-2\right)^5\)

câu d thì dễ

Thế giúp vs bn ơi

mk bn lớp mấy

(3¹⁵.4+3¹⁴.15):3¹⁶

=(3¹⁵.4+3¹⁴.3.5):3¹⁶

=(3¹⁵.4+3¹⁵.5):3¹⁶

=[3¹⁵.(4+5)]:3¹⁶

=[3¹⁵.9]:3¹⁶

=3¹⁵.3²:3¹⁶

⁼³

(1+2+3+4+....+1000)x(512-512)

=(1+2+3+4+....+1000)X0=0

thanks bn nha mờ chỉ viết vậy thui àk Nguyễn Tiến Dũng

Ta có:a)\(^{3^{600}}\)=\(^{\left(3^3\right)^{200}}\)=\(^{27^{200}}\)                                                 \(^{4^{400}}\)=\(^{\left(4^2\right)^{200}}\)=\(^{16^{200}}\)

vì 27^200>16^200             =>   3^600>4^400

b)   \(^{4^{32}=4^{2.16}=16^{16}}\)                 vì 16^16>16^15      =>   4^32>16^15

\(3^{600}=3^{200.3}=\left(3^3\right)^{200}=9^{200}^{_{\left(1\right)}}\)

\(4^{400}=\left(2^2\right)^{400}=2^{800}=2^{200.4}=\left(2^4\right)^{200}=16^{200}_{\left(2\right)}.\)

\(\left(1\right),\left(2\right)\Rightarrow4^{400}>3^{600}\)

\(4^{32}=\left(2^2\right)^{32}=2^{64}_{\left(1\right)}\)

\(16^{15}=\left(2^4\right)^{15}=2^{60}_{\left(2\right)}\)

\(\left(1\right),\left(2\right)\Rightarrow4^{32}>16^{15}\)