a) B = 124 x 122 = (123+1) x (123-1) = 123 x 123 -123 + 123 -1 = A -1

=> B < A

b) B = 986 x 985 = (987-1) x (984+1) = 987 x 984 + 987 - 984 -1 = A +2

=> B > A

a) Ta có: A = 123 x 123 = 123 x ( 124 - 1 ) = 123 x 124 - 123
               B = 121 x 124 = ( 123 - 2 ) x 124 = 123 x 124 - 248
Suy ra: 123 x 124 - 123 > 123 x 124 - 248 hay A > B

b) Ta có: C = 123 x 137137 = 123 x 137 x 1001 = 137 x 123 x 1001
               D = 137 x 123123 = 137 x 123 x 1001
Suy ra: C = D

c) Ta có: E = 2015 x 2017 = 2015 x ( 2016 + 1 ) = 2015 x 2016 + 2015
               F = 2016 x 2016 = ( 2015 + 1 ) x 2016 = 2015 x 2016 + 2016
Suy ra: 2015 x 2016 + 2015 < 2015 x 2016 + 2016 hay E < F 
 

a) `123 - 27 + 27 + 13 - 123`

`= (123 - 123) + (27 - 27) + 13`

`= 0 + 0 + 13`

`= 13`

`---`

b) `175 + 25 + 13 + 15 - 175 - 25`

`= (175-175)+(25-25)+25+13`

`=0+0+25+13`

`=28`

`---`

c) \(53\times39+47\times39+53\times21+47\times21\)

\(=\text{[}\left(53+47\right)\times39]+\left[\left(53+47\right)\times21\right]\)

\(=\left(100\times39\right)+\left(100\times21\right)\)

\(=3900+2100\)

`=6000`

`---`

d) \(49\times35+49\times16-49\times61\)

`=` \(49\times\left(35+16-61\right)\)

`=` \(49\times\left(-10\right)\)

`=-490`

A = \(\dfrac{3^{123}+1}{3^{125}+1}\)  Vì 3¹²³ + 1 < 2¹²⁵ + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)

A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B

Vậy A < B 

Ta có : \(\frac{124124}{125125}=\frac{124124:1001}{125125:1001}=\frac{124}{125}\)

\(\frac{123}{124}=1-\frac{1}{124}\); \(\frac{124}{125}=1-\frac{1}{125}\)

Vì 1/124 < 1/125 => 123/124 > 123/125 => 123/124 > 124124/125125

ta có: \(\frac{124124}{125125}=\frac{124}{125}\)

Lại có: \(1-\frac{123}{124}=\frac{1}{124};1-\frac{124}{125}=\frac{1}{125}\)

\(\Rightarrow\frac{1}{124}>\frac{1}{125}\Rightarrow1-\frac{123}{124}>1-\frac{124}{125}\)

\(\Rightarrow\frac{123}{124}< \frac{124}{125}\Rightarrow\frac{123}{124}< \frac{124124}{125125}\)

a)

\(a=2136\times2136=2136\times\left(2134+2\right)=2136\times2134+2136\times2\)

\(b=2134\times2138=2134\times\left(2136+2\right)=2134\times2136+2134\times2\)

Vì \(2136\times2>2134\times2\)     

=> \(a>b\)

b)

=> \(a=b\)\(b=123\times246+77-=123\times\left(245+1\right)+77=123\times245+123+77=123\times245+200\)\(a=245\times124-45=245\times\left(123+1\right)-45=245\times123+245-45=245\times123+200\)

                                                               Bài giải

a, Ta có : \(121\cdot124=121\cdot123+121\)

                 \(123\cdot123=121\cdot123+2\cdot123=121\cdot123+246\)

Vì \(121\cdot123+246>121\cdot123+121\) nên \(123\cdot123>121\cdot124\)

b, Ta có : 

\(2004\cdot2004=2002\cdot2004+2\cdot2004=2002\cdot2004+4008\)

\(2002\cdot2006=2002\cdot2004+2\cdot2002=2002\cdot2004+4004\)

Vì \(2002\cdot2004+4008>2002\cdot2004+4004\) nên \(2004\cdot2004>2002\cdot2006\)

Ta có :

S = 1 21 + 1 22 + 1 23 + ⋯ + 1 29 + 1 30 > 1 30 + ⋯ 1 30 = 10 30 = 1 3 .

Vậy S >  1 3