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

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

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

S = 1 x 2 + 2 x 3 + 3 x 4 + ... + 38 x 39 + 39 x 40

3S = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 38 x 39 x 3 + 39 x 40 x 3

3S = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ... + 38 x 39 x ( 40 - 37 ) + 39 x 40 x ( 41 - 38 )

3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 38 x 39 x 40 - 37 x 38 x 39 + 39 x 40 x 41 - 38 x 39 x 40

S = 39 x 40 x 41 : 3

S = 21320

S=1x2+2x3+3x4+...+38x39+39x40

3S=(1x2+2x3+3x4+...+38x39+39x40)x3

3S=1x2x3+2x3x3+3x4x3+....+39x40x3

3S=1x2x3+2x3x(4-1)+3x4x(5-2)+...+39x40x(41-38)

3S=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+...+39x40x41-38x39x40

3S=39x40x41

3S=63960

S=63960:3

S=21320

202^203<203^202

Ta có :

202²⁰³ = 8 242 408¹⁰¹ ( 1 )

203²⁰² = 42 209¹⁰¹       ( 2 )

Từ ( 1 ) và ( 2 ) suy ra 202²⁰³ < 203²⁰²

vì cơ số 202<cơ số 203 nên 202^203<203^202

Nếu là violympic

thì cần kq thôi hả

S = 21320

S = 1.2 + 2.3 + 3.4 + ... + 38.39 + 39.40

3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 38.39.(40-37) + 39.40.(41-38)

3S = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 38.39.40 - 37.38.39 + 39.40.41 - 38.39.40

3S = 39.40.41

S = 13.40.41

S = 21320

S = 1x2+2x3+3x4+ ... + 38x39 + 39 x 40

3S = 1x2x3 + 2x3x3 + 3x4x3 + ... + 38x39x3 + 39x40x3

3S = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + ... + 38x39x( 40-37 ) + 39x40x(41-38)

3S = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + ... + 38x39x40 - 37x38x39 + 39x40x41 - 38x39x40

3S = 39x40x41

S = 39x40x41 : 3

S = 21320

S = 1 x 2 + 2 x 3 + 3 x 4 + ... + 38 x 39 + 39 x 40

3S = 1 x 2 x ( 3 - 0 ) + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + ... + 38 x 39 x ( 40 - 37 ) + 39 x 40 x ( 41 - 38 )

3S = 1 x 2 x 3 - 0 x 1 x 2 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 38 x 39 x 40 - 37 x 38 x 39 + 39 x 40 x 41 - 38 x 39 x 40

S = \(\frac{39x40x41}{3}\)

S = 21320

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