mik cần gấp

Xét B =  \(\frac{201+202+203}{202+203+204}\)

          = \(\frac{201}{202+203+204}\)+ \(\frac{202}{202+203+204}\)+ \(\frac{203}{202+203+204}\)

 Vì 202 < 202 + 203 + 204 nên \(\frac{201}{202}\)>\(\frac{201}{202+203+204}\)(1)

Vì 203 < 202 + 203 + 204 nên \(\frac{202}{203}\)> \(\frac{202}{202+203+204}\)(2)

Vì 204 < 202 + 203 + 204 nên \(\frac{202}{203}\)>\(\frac{202}{202+203+204}\)(3)

Cộng vế vơi vế của (1) , (2) và (3)

=>\(\frac{201}{202}+\frac{202}{203}+\frac{203}{204}\)> \(\frac{201+202+203}{202+203+204}\)

=> A > B

Vậy A > B

Xét B = \(\frac{201+202+203}{202+203+204}\)

         = \(\frac{201}{202+203+204}\)\(+\)\(\frac{202}{202+203+204}\)\(+\)\(\frac{203}{202+203+204}\)

Vì 202 < 202 + 203 + 204

=> \(\frac{201}{202}\)> \(\frac{201}{202+203+204}\)( 1 )

Vì 203 < 202 + 203 + 204

=> \(\frac{202}{203}\)>\(\frac{202}{202+203+204}\)( 2 )

Vì 204 < 202 + 203 + 204

=> \(\frac{203}{204}\)> \(\frac{203}{202+203+204}\)( 3 )
Cộng vế với vế của ( 1 ), ( 2 ) và ( 3 )

=> \(\frac{201}{202}+\frac{202}{203}+\frac{203}{204}\)> \(\frac{201+202+203}{202+203+204}\)

=> A > B

Vậy A > B

..................tên em là jullei Trinh...........................

Ta có : \(202^{203}=(2\cdot101)^{3\cdot101}=(1^3\cdot101^3)^{101}=(8\cdot101\cdot10^{12}\cdot101)=(808\cdot1012)^{101}\)

           \(303^{202}=(3\cdot101)^{2\cdot101}=(32\cdot101^2)^{101}=(9\cdot101^2)^{101}\)

\(\Rightarrow(808\cdot101^2)>(9\cdot101^2)\)

Vậy : 

202³⁰³=(2.101)^3.101=(2³.101³)¹⁰¹=(8.101.101²)¹⁰¹=(808.101²)¹⁰¹

303²⁰²=(3.101)^2.101=(3².101²)¹⁰¹=(9.101²)¹⁰¹

Vì (808.101²)¹⁰¹ > (9.101²)¹⁰¹ suy ra 202³⁰³> 303²⁰²

Vậy.......

a, 202²⁰³=(101.2)²⁰³

=101²⁰³.2²⁰³

=101²⁰².2²⁰².202

b, 203²⁰²=(101,5.2)²⁰²

=101,5²⁰².2²⁰²

còn lại dễ

b, 1990¹⁰+1990⁹=1990⁹.1990+1990⁹=1990⁹.(1990+1)=1990⁹.1991

1991²⁰=1991¹⁹.1991

=>1990¹⁰+1990⁹<1991²⁰

c, 11¹⁹⁷⁹<11¹⁹⁸⁰=(11³)⁶⁶⁰=1331⁶⁶⁰

37¹³²⁰=(37²)⁶⁶⁰=1369⁶⁶⁰

=>11¹⁹⁷⁹<37¹³²⁰

a= 202 x 204 = 202 x (203+1)=202 x 203 + 202

b=203 x 203 = (202+1) x 203 = 202 x 203 + 203

Vì 203>202 => 202x 203 + 202<202x203 +203

=>a<b

a=(203-1)(203+1)=203^2-1<203^2=b

\(202^{303}=\left(2.101\right)^{3.101}=\left(2^3.101^3\right)^{101}=\left(8.101^3\right)^{101}\)

\(303^{202}=\left(3.101\right)^{2.101}=\left(3^2.101^2\right)^{101}=\left(9.101^2\right)^{101}\)

Mà \(8.101^3>9.101^2\)

\(\Rightarrow202^{303}>303^{202}\)

\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)

Nhận thầy 10⁸ - 1 > 10⁸ - 3

=> \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

=> \(1+\frac{3}{10^8-1}< \frac{3}{10^8-3}+1\)

=> A < B

b) 17C = \(\frac{17\left(17^{203}+1\right)}{17^{204}+1}=\frac{17^{204}+1+16}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)

17D = \(\frac{17\left(17^{202}+1\right)}{17^{203}+1}=\frac{17^{203}+1+16}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)

Nhận thầy 17²⁰³ + 1 < 17²⁰⁴ + 1

=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)

=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\) 

cho mình 1 tích đi mình hộ nha k thì thoi

202^303 và 303^202 
202^(3.101) và 303^(2.101) 
(202^3)^101 và (303^2)^101 
202^3 và 303^2 
(2.101)^3 va (3.101)^2 
2^3.101^3 va 3^2.101^2 
8.101.101^2 va 9.101^2 
8.101 va 9 
808 > 9 => 202^303 > 303^202

203³⁰³=( 203³)¹⁰¹ = 8365427¹⁰¹

303²⁰²=( 303²)¹⁰¹ = 91809¹⁰¹

=> 203³⁰³ > 303²⁰²

tk mình nha