Bài 1 : Viết các tổng sau thành bình phương của 1 số tự nhiên 
A. 5 ³ + 6² + 8
B . 2 + 3²+ 4² + 13²

Bài 2 : So sánh các số sau 
 A . 3²⁰ và 27⁴

Ta có : 27⁴ = (3²)⁴ = 3⁸ 

Vì 20 < 8 => 3²⁰ > 27⁴

( Những câu còn lại tương tự ) - Tự làm nhé ! Mình bận ~

# Dương 

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

Trả lời:

a, Ta có: 3²⁰ ; 27⁴ = ( 3³ )⁴ = 3¹²

Vì 3²⁰ > 3¹² nên 3²⁰ > 27⁴

b, 2²⁵ ; 16⁶ = ( 2⁴ )⁶ = 2²⁴

Vì 2²⁵ > 2²⁴ nên 2²⁵ > 16⁶ 

Trả lời:

c, 10³⁰ = ( 10³ )¹⁰  = 1000¹⁰ ; 4⁵⁰ = ( 4⁵ )¹⁰ = 1024¹⁰

Vì 1000¹⁰ < 1024¹⁰ nên 10³⁰ < 4⁵⁰

d, 5³⁴ ; 25.5³⁰ = 5² . 5³⁰ = 5³²

Vì 5³⁴ > 5³² nên 5³⁴ > 25.5³⁰

`#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) Ta có :

\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)

\(\Rightarrow27^{27}>243^{13}\)

\(\Rightarrow-27^{27}< -243^{13}\)

\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)

b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)

c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)

\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)

\(\Rightarrow4^{50}>8^{30}\)

d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)

\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)

a) Ta có :

$27^{27}>27^{26}={\left(27^2\right)}^{13}=729^{13}>243^{13}$

$\Rightarrow 27^{27}>243^{13}$

$\Rightarrow -27^{27}<-243^{13}$

$\Rightarrow {\left(-27\right)}^{27}<{\left(-243\right)}^{13}$

b) ${\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{26}={\left(\genfrac{}{}{0ex}{}{1}{8^2}\right)}^{13}={\left(\genfrac{}{}{0ex}{}{1}{64}\right)}^{13}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(-\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}<{\left(-\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

c) $4^{50}={\left(4^5\right)}^{10}=1024^{10}$

$8^{30}={\left(8^3\right)}^{10}=512^{10}<1024^{10}$

$\Rightarrow 4^{50}>8^{30}$

d) ${\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{12}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

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