9²⁰⁰ = (9²)¹⁰⁰ = 81¹⁰⁰ < 99¹⁰⁰ .Vậy 9²⁰⁰ < 99¹⁰⁰

ta có : \(9^{200}=9^{2.100}=\left(9^2\right)^{100}=81^{100}\)

           vì \(81< 99\Rightarrow81^{100}< 99^{100}\)

                               \(\Rightarrow9^{200}< 99^{100}\)

9²⁰⁰ = 9^2.100 = ( 9² ) ¹⁰⁰ = 81¹⁰⁰

99¹⁰⁰ = 99¹⁰⁰

Vì 81 < 99 => 9²⁰⁰ < 99¹⁰⁰

\(99^{100}=99^{100}\)

\(33^{200}=\left(33^2\right)^{100}=1089^{100}\)

vi \(1089>99\)  nen \(1089^{100}>99^{100}\)

vay \(99^{100}< 33^{200}\)

99^{100 va} 33²⁰⁰

99^{100 <} 33²⁰⁰

ta có : 9^200=(9^2)^100=81^100.

vì 81^100<99^100 nên 9^200<99^100.

\(9^{200}=\left(9^2\right)^{100}=81^{100}< 99^{100}\)

Vậy  \(9^{200}< 99^{100}\)

\(9^2^{00}=\left(9^2\right)^{100}=81^{100}\)

\(99^{100}=\left(99^1\right)^{100}=99^{100}\)

\(81< 99=>81^{100}< 99^{100}\)

\(9^{200}< 99^{100}\)

ta có 9999= 99 *101. 
do đó 9999^10 = 99 ^10 * 101^10 
còn 99^20 = 99^10 * 99^10 
vì 99^10 < 101^10 nên 99^10 * 99^10 < 99 ^10 * 101^10 . 
vậy 99^20 < 9999^10. 

ta co:9999^10 = 99^10 x 99^10 x 2^10 = 99^200

        suy ra :9999^10=99^200

vay . . .

 TA CÓ : 

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

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

hok tốt~~

99²⁰<9999¹⁰.

Ta có : 99²⁰ = 99^{2 . 10} = 9801¹⁰

   => 9801¹⁰ < 9999¹⁰ nên 99²⁰ < 9999¹⁰

`A=3/4+8/9+.............+9999/10000`

`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`

`=99-(1/4+1/9+.........+1/10000)<99-0=99`

`=>A<99`

 địt mẹ con ngu t khinh

Giải:

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\) 

\(A=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{8}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{10000}\right)\) 

\(A=99-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{19}+...+\dfrac{1}{10000}\right)< 99\) 

\(\Rightarrow A< 99\left(đpcm\right)\) 

Chúc bạn học tốt!

a/ Ta có

\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)

\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)

\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)

\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)

Thế lại bài toán ta được:

\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)

\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)

b/ Ta có: 

A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)

\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)

Vậy A < B

cảm ơn bạn

\(2^{50}=\left(2^5\right)^{10}=32^{10}\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}\)

Suy ra: 2⁵⁰ > 5²⁰

b)

\(9^{200}=\left(9^2\right)^{100}=81^{100}\)

Suy ra: 99¹⁰⁰ > 81¹⁰⁰

\(5^{202}=\left(5^2\right)^{101}=25^{101}\)

\(2^{505}=\left(2^5\right)^{101}=32^{101}\)

Suy ra: 5²⁰² < 2⁵⁰⁵