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,?????

Lời giải:

$78^{15}-78^{12}=78^{12}(78^3-1)> 78^9(78^3-1)=78^{12}-78^9$

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

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

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

2 mũ 1000 > 12 mũ 1200

9 mũ 99 > 99 mũ 9

a, 12^1200 > 2^100 Vì cả cơ số lẫn số mũ đều lớn hơn

b, 9^99= (9^11)^9

Vì 9^11> 99 nêm 99^11^9> 99^9

Vậy 9^99> 99^9