\(2^{225}=8^{75}< 9^{75}=3^{150}\)

\(2^{91}>2^{90}=32^{18}>25^{18}=5^{36}>5^{35}\)

\(99^{20}=\left(99.99\right)^{10}< \left(99.101\right)^{10}=9999^{10}\)

a, \(2^{225}=\left(2^3\right)^{75}\) 

    \(3^{150}=\left(3^2\right)^{75}\)

b,\(2^{91}=\left(2^{13}\right)^7\)

\(5^{35}=\left(5^5\right)^7\)

c,\(99^{20}=\left(99\cdot99\right)^{10}\)

\(9999^{10}=\left(99\cdot101\right)^{10}\)

Khỏi cần nữa rồi...Mới nghĩ ra rồi .-.

Mà tôi không chơi kiểu tick đã rồi trả lời.Ok ?

Ta co : 99²⁰=99^2.10=(99²)¹⁰=9801¹⁰

9999¹⁰⁼9999^1.10=(9999¹)¹⁰=9999¹⁰

Vi 9801¹⁰<9999¹⁰=>99²⁰<9999¹⁰

TICK cho minh nha!!! 

Lời giải:

a. $\tan 25^0=\frac{\sin 25^0}{\cos 25^0}> \sin 25^0$ do $0< \cos 25^0< 1$

b. $\cot 32^0 = \frac{\cos 32^0}{\sin 32^0}> \cos 32^0$ do $0< \sin 32^0< 1$
c. $\tan 45^0= 1; \cos 45^0=\frac{\sqrt{2}}{2}$ nên $\tan 45^0> \cos 45^0$

d. $\cot 60^0= \frac{\cos 60^0}{\sin 60^0}=\frac{\sin 30^0}{\sin 60^0}> \sin 30^0$ do $0< \sin 60^0< 1$

 TA CÓ : 

\(99^{20}=99^{2\times10}=9801\)

\(\Rightarrow9801^{10}< 9999^{10}\) nên \(99^{20}< 9999^{10}\)

hok tốt~~

99²⁰<9999¹⁰.

99²⁰ =(99²)¹⁰=(99x99)¹⁰<(99x101)¹⁰=9999¹⁰ 

vậy 99²⁰²⁰ <9999¹⁰

2.B=1+5+5^2+...+5^98

B=1+5^2+5^3+...+5^96+5^97+5^98

B=(1+5+5^2)+(5^3+5^4+5^5)+...+(5^96+5^97+5^98)
B=(1+5+25)+5^3.(1+5+25)+...+5^96.(1+5+25)

B=31+5^3.31`+...+5^96.31

B=(1+5^3+...+5^98).31.Suy ra B chia hết cho 31.

\(2^{91}=\left(2^{13}\right)^7=73728^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)

\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)

\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)

Ta có : \(2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{222}=\left(3^2\right)^{111}=9^{111}\)

Do : \(8^{111}< 9^{111}\left(8< 9\right)\)

\(\Rightarrow2^{333}< 3^{222}\)

Ta có : \(9^{1005}=\left(3^2\right)^{1005}=3^{2010}\)

Do : \(3^{2009}< 3^{2010}\left(2009< 2010\right)\)

\(\Rightarrow3^{2009}< 9^{1005}\)