a) \(21^{15}=21^{3.5}=\left(21^3\right)^5=9261^5\)

Vì \(9261>27\Rightarrow9261^5>27^5\Rightarrow21^{15}>27^5\)

b) \(15^{12}=\left(3.5\right)^{12}=3^{12}.5^{12}\)

\(81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{4.3}.5^{3.5}=3^{12}.5^{15}\)

Vì \(3^{12}=3^{12}\)mà \(5^{12}< 5^{15}\Rightarrow3^{12}.5^{12}< 3^{12}.5^{15}\Rightarrow15^{12}< 81^3.125^5\)

199 ²⁰  < 2003¹⁵

3³⁹ < 11²¹

tớ nghĩ là vậy . 

tại tớ không tính 

a) 5²³ và 6.5²² 
Ta có:  5.5²² = 5²³
Mà: 5.5²² < 6.5²² 
=> 5²³< 6.5²²

b) 21¹⁵ và 27⁵. 49⁸

Ta có: 21¹⁵= 3¹⁵ . 7¹⁵

27⁵ .49⁸ = 3^(3.5) . 7^(2.8)=3¹⁵.7¹⁶

Mà:  7¹⁵ < 7¹⁶

=> 21¹⁵ < 27⁵. 49⁸

c) 199^{20 va} 2003¹⁵

Ta có:   2003¹⁵ > 2000¹⁵ = (2.10³)¹⁵ = 2¹⁵.10⁴⁵ 
           199²⁰ < 200²⁰ = (2.10²)^{^20} = 2²⁰.10⁴⁰ =2¹⁵.2⁵.10⁴⁰ < 2¹⁵.10⁴⁰.100 =2¹⁵.10⁴² 
Mà 10⁴⁵ > 10⁴² 

=> 199²⁰  < 2003¹⁵

d) 3³⁹ và 11²¹

Ta có:  3³⁹ < 3⁴⁰ = 9²⁰ < 11²⁰

Mà  11²⁰ < 11²¹

=> 3³⁹ < 11²¹

199 ²⁰ < 2003¹⁵ 

3³⁹ < 11²¹ 

nhé bạn !

Đ/s : 2 dấu lớn

tất cả đều đâu <

a/ 

\(37^{1320}=\left(37^2\right)^{660}=1369^{660}\)

\(11^{1979}< 11^{1980}=\left(11^3\right)^{660}=1331^{660}\)

\(\Rightarrow1363^{660}>1331^{660}\Rightarrow37^{1320}>11^{1979}\)

b/

\(27^{11}=\left(3^3\right)^{11}=3^{33}\)

\(81^8=\left(3^4\right)^8=3^{32}\)

\(\Rightarrow27^{11}>81^8\)

d/

\(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 9^{21}< 11^{21}\)

e/ \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

\(11^{24}=\left(11^2\right)^{12}=121^{12}\)

\(\Rightarrow5^{36}>11^{24}\)

g/ \(21^{15}=3^{15}.7^{15}\)

\(27.49^8=3^3.\left(7^2\right)^8=3^3.7^{16}\)

\(\frac{21^{15}}{27.49^8}=\frac{3^{15}.7^{15}}{3^3.7^{16}}=\frac{3^{12}}{7}>1\Rightarrow21^{15}>27.49^8\)

f/ \(199^{20}=\left(199^4\right)^5\)

\(2003^{15}=\left(2003^3\right)^5\)

\(2003^5>1990^5\)

\(\frac{1990^5}{199^4}=\frac{199^5.10^5}{199^4}=199.10^5>1\)

\(\Rightarrow2003^5>1990^5>199^4\Rightarrow2003^{15}>199^{20}\) 

a, 9^5>27^3

​b,3^200>2^300

​c, 32^11<17^14

37^1320=(37^2)^660=1369^660