B = \(\frac{7}{2.4}+\frac{7}{4.6}+\frac{7}{6.8}+.....+\frac{7}{100.102}\)

ta có \(\frac{7}{2.4}=\frac{1}{2}\left(\frac{7}{2}-\frac{7}{4}\right);\frac{7}{4.6}=\frac{1}{2}\left(\frac{7}{4}-\frac{7}{8}\right);......;\frac{7}{100.102}=\frac{1}{2}\left(\frac{7}{100}-\frac{7}{102}\right)\)

⇒ B = \(\frac{1}{2}\left(\frac{7}{2}-\frac{7}{4}+\frac{7}{4}-\frac{7}{8}+....+\frac{7}{100}-\frac{7}{102}\right)\)

⇔ B = \(\frac{1}{2}\left(\frac{7}{2}-\frac{7}{102}\right)\)

⇔ B = \(\frac{1}{2}.\frac{175}{51}\)

⇔ B = \(\frac{175}{102}\)

A = \(\dfrac{47\times48-47\times47-24-23+2046}{2+4+8+..+512+1024}\)

Đặt tử số là B, mẫu số là C thì

B = 47\(\times\)48 - 47 \(\times\) 47 - 24 -23+2046vàC = 2 + 4 + 8 +....+ 512 + 1024

B = 47 \(\times\) 48 - 47 \(\times\) 47 - 24 - 23 + 2046

B= 47 \(\times\) 48 - 47 \(\times\) 47 - ( 24 + 23) + 2046

B = 47 \(\times\) 48 - 47 \(\times\) 47 - 47 + 2046

B = 47 \(\times\) 48 - 47 \(\times\) 47 - 47 \(\times\) 1 + 2046

B = 47 \(\times\) ( 48 - 47 - 1) + 2046

B = 47 \(\times\) 0 + 2046

B = 2046 

C                =    2 + 4 + 8+ ....+ 512 +1024

C \(\times\) 2        =            4 + 8 +.....+ 512 + 1024 + 2048

C \(\times\) 2 - C   =          2048 - 2 

C \(\times\) ( 2 - 1) =          2046

C                 = 2046

A = \(\dfrac{B}{C}\) = \(\dfrac{2046}{2046}\) = 1

Ta có :

\(\dfrac{3}{2}\) = \(\dfrac{3}{1.2}\) = 3 x \(\dfrac{1}{1.2}\) = 3 x ( 1 - \(\dfrac{1}{2}\) ) : 2 ;

\(\dfrac{3}{8}\) = \(\dfrac{3}{2.4}\) = \(\)3 x \(\dfrac{1}{2.4}\) = 3 x ( \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ) : 2 ;

\(\dfrac{3}{24}\) = \(\dfrac{3}{4.6}\) = 3 x \(\dfrac{1}{4.6}\) = 3. ( \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) ) : 2 ;

\(\dfrac{3}{9800}\) = \(\dfrac{3}{98.100}\) = 3 x \(\dfrac{1}{98.100}\) = 3 x ( \(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) ) : 2 ;

\(\dfrac{3}{10200}\) = \(\dfrac{3}{100.102}\) = 3 x \(\dfrac{1}{100.102}\)= 3 x ( \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) )

Vậy : \(\dfrac{3}{2}\) + \(\dfrac{3}{8}\) + \(\dfrac{3}{24}\) + ........ + \(\dfrac{3}{9800}\) + \(\dfrac{3}{10200}\)

= \(\dfrac{3}{1.2}\) + \(\dfrac{3}{2.4}\) + \(\dfrac{3}{4.6}\) +........+ \(\dfrac{3}{98.100}\) + \(\dfrac{3}{100.102}\)

= 3 x ( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + .........+ \(\dfrac{1}{98.100}\) + \(\dfrac{1}{100.102}\) )

= 3 x ( \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\) +.................+ \(\dfrac{2}{98.100}\) + \(\dfrac{2}{100.102}\) ) : 2

= 3 x ( 1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{6}\) +.......+\(\dfrac{1}{98}\) - \(\dfrac{1}{100}\) + \(\dfrac{1}{100}\) - \(\dfrac{1}{102}\) ) : 2

= 3 x ( 1 - \(\dfrac{1}{102}\) ) : 2 = 3 x \(\dfrac{101}{102}\) : 2

= \(\dfrac{101}{68}\)

Ta có: \(A=\dfrac{5}{8}+\dfrac{5}{24}+\dfrac{5}{48}+...+\dfrac{5}{9800}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{8}+\dfrac{2}{24}+\dfrac{2}{48}+...+\dfrac{2}{9800}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{49}{50}\)

\(=\dfrac{245}{100}=\dfrac{49}{20}\)

B = \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\) + ..... + \(\frac{2}{99.101}\)+ \(\frac{2}{101.103}\)

 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/99 - 1/101 + 1/101 - 1/103

= 1- 1/103 = 102/103

a)A=(-123) - 77 + (-257) +23 - 43                                                         b)B=48+| 48-174|+(-74)

   A=[(-123) - 77]+[(-257)-43]+23                                                              B=48+(174-48)+(-74)

  A= -200+(-300)+23                                                                                B=48+174+(-48)+(-74)

  A= -500+23                                                                                            B=[48+(-48)]+[174+(-74)]

  A= -477                                                                                                   B=0+100=100

c)C= -2012+(-596)+(-201)+496+301                                                      d)D=1+2-3-4+5+6-7-8+............-79-80-81  

   C= -2012+[(-596)+496]+[(-201)+301]                                                     D=1+(2-3-4+5)+(6-7-8+9)+............+(78-79-80-81)

   C= -2010+(-100)+100                                                                             D=1+0+0+............+(-162)

   C= -2010+0                                                                                            D=1+(-162)

   C= -2010                                                                                                 D= -161