Trả lời

4¹⁰⁰=2²⁰⁰

2²⁰²

Vậy 2²⁰⁰ < 2²⁰² hay 4¹⁰⁰ < 2²⁰²

3⁰ và 5⁸

3⁰ < 5⁸

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d, 

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

a,3¹² và 5⁸

Ta có:3¹²=(3³)⁴=27⁴

5⁸=(5²)⁴=25⁴

Vì 27⁴>25⁴ nên 3¹²>5⁸

b,(0,6)⁹ và (0,9)⁶

Ta có:(0,9)⁶>(0,6)⁶ mà (0,6)⁶>(0,6)⁹

\(\Rightarrow\)(0,6)⁹<(0,9)⁶

c,5²⁰⁰⁰ và 10¹⁰⁰⁰

Ta có:5²⁰⁰⁰=(5²)¹⁰⁰⁰=25¹⁰⁰⁰>10¹⁰⁰⁰

\(\Rightarrow\)5²⁰⁰⁰>10¹⁰⁰⁰

d,?????

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

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

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

\(6^8và16^{12}=\left(6.8\right)^0và\left(16.3\right)^9=48< 48^9\)

6⁸ = (6²)⁴ = 36⁴

16¹² = (16³)⁴ = 4096⁴

Do 36 < 4096 nên 36⁴ < 4096⁴

Vậy 6⁸ < 16¹²

Ta có:\(8^{12}=\left(2^3\right)^{12}=2^{3.12}=2^{36}\\ \\ \\ 32^6=\left(2^5\right)^6=2^{5.6}=2^{30}\) Mà \(2^{36}>2^{30}\)

⇒ Chọn A

8¹² = (2³)¹² = 2³⁶

32⁶ = (2⁵)⁶ = 2³⁰

Vì 2³⁶ > 3³⁰ ⇒ 8¹² > 32⁶ ⇒ Chọn A

Ta có : 12⁸ . 9¹² = ( 3. 4 )⁸. (3²)¹²

= 3⁸. 4⁸. 3²⁴

= (2²)⁸. 3³²

=2¹⁶. (3²)¹⁶

= 2¹⁶. 9¹⁶

= (2 .  9)¹⁶

=18¹⁶

ý thứ hai làm tương tự

64⁸=(2⁶)⁸=2⁴⁸

16¹²=(2⁴)¹²=2⁴⁸

Vì 2⁴⁸=2⁴⁸ nên 64⁸=16¹²

64⁸ = (2⁶)⁸ = 2⁴⁸

16^{12 =} (2⁴)¹² = 2⁴⁸

Vì 2⁴⁸ = 2⁴⁸ => 16¹² = 64⁸

Vậy 64⁸ = 16¹²