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

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

\(\Rightarrow3S=1x2\left(3-0\right)2x3\left(4-1\right)+...+39x40\left(41-38\right)\)

\(\Rightarrow3S=1x2x3-0x1x2+2x3x4-1x2x3+...+39x40x41-38x39x40\)

\(\Rightarrow S=\frac{38x39x40}{3}\)

\(\Rightarrow S=21320\)

\(S=1.2+2.3+3.4+...+38.39+39.40\)

\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+38.39.\left(40-37\right)+39.40.\left(41-38\right)\) 

\(3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+38.39.40-37.38.39+39.40.41-38.39.40\)

\(3S=39.40.41\)

\(S=\frac{63960}{3}=21320\)

. là nhân

S =  1x2+2x3+3x4+...+39x40

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

S = 39x40x41 : 3

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 + 3.4.5 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 38.39.40 - 37.38.39 + 39.40.41 - 38.39.40

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

3S = 39.40.41 - 0.1.2

3S = 39.40.41

S   = 39.40.41 : 3

S   = 21320

phép tính đâu =???????

các bạn chỉ cần trả lời đáp án là bao nhiêu thôi 

ai trả lời nhanh và đung thì mình sẽ tích cho

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

<=>3S= 1.2.3 +2.3.3+ 3.4.3 +...+ 38.39.3 + 39.40.3

<=>3S= 1.2.3 + 2.3.(4-1) + 3.4.(5-2) +...+ 38.39.(40-37)+ 39.40.(41-38)

<=>3S= 1.2.3 + 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= \(\dfrac{39.40.41}{3}\)

<=>S=21320

\(S=1\cdot2+2\cdot3+3\cdot4+...+39\cdot40\)

\(3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+39\cdot40\cdot\left(41-38\right)\)

\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+39\cdot40\cdot41-38\cdot39\cdot40\)

\(3S=39\cdot40\cdot41\)\(\Rightarrow S=\dfrac{39\cdot40\cdot41}{3}=21320\)

Ta có : \(S=1.2+2.3+3.4+...+38.39+39.40\)

\(3S=1.2.3+2.3.3+3.4.3+...+38.39.3+39.40.3\)

\(3S=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+39.40.\left(41-38\right)\) \(3S=39.40.41\)

\(S=\dfrac{39.40.41}{3}\)\(=21320\)

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

em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

       D =           \(\dfrac{1}{7^2}\) - \(\dfrac{2}{7^3}\) + \(\dfrac{3}{7^4}\) - \(\dfrac{4}{7^5}\) +........+ \(\dfrac{201}{7^{202}}\) -  \(\dfrac{202}{7^{203}}\)

7 \(\times\) D  =  \(\dfrac{1}{7}\) -  \(\dfrac{2}{7^2}\) +  \(\dfrac{3}{7^3}\) - \(\dfrac{4}{7^4}\)  + \(\dfrac{5}{7^5}\) -.......- \(\dfrac{202}{7^{202}}\)

7D +D  =   \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)

         D = (  \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -.........-\(\dfrac{1}{7^{202}}\) - \(\dfrac{202}{7^{203}}\)) : 8

Đặt    B =      \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^4}\) + \(\dfrac{1}{7^5}\) -........+\(\dfrac{1}{7^{201}}\).-\(\dfrac{1}{7^{202}}\) 

  7   \(\times\) B = 1 - \(\dfrac{1}{7}\)+\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) + \(\dfrac{1}{7^4}\) - \(\dfrac{1}{7^5}\) +.........- \(\dfrac{1}{7^{201}}\)

7B + B   =  1 - \(\dfrac{1}{7^{202}}\)

          B   =  ( 1 - \(\dfrac{1}{7^{202}}\)) : 8

         D  =  [ ( 1 - \(\dfrac{1}{7^{202}}\)): 8  - \(\dfrac{202}{7^{203}}\)] : 8 

          D = \(\dfrac{1}{64}\) - \(\dfrac{1}{64.7^{202}}\) - \(\dfrac{202}{7^{203}.8}\) < \(\dfrac{1}{64}\)

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