Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)

= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\) 

B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\) 

Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B

Vậy A < B

Cảm ơn bạn nhé!

Bài 1:

Ta có:

\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)

\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)

\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+...+\left(1+\dfrac{48}{2}\right)+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)

Bài 2:

Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\)

Nhận xét:

\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{2}\)

1: =29-16-3*2=13-6=7

2: =35-12+(-14)-2

=23-16=7

3: =4*3125-32:16

=12500-2

=12498

4: =-452+67-75+452=-8

5: =18+99*18-3^5-2^5

=1800-275

=1525

= \(\dfrac{49}{2}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}\)

= \(\dfrac{784}{32}-\dfrac{16}{32}-\dfrac{8}{32}-\dfrac{4}{32}-\dfrac{2}{32}-\dfrac{1}{32}\)

= \(\dfrac{753}{32}\)

       A =       \(\dfrac{49}{2}\) - \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\) - \(\dfrac{1}{32}\)

A \(\times\) 2  = 49 - 1 - \(\dfrac{1}{2}\)  - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) - \(\dfrac{1}{16}\)

A \(\times\) 2  - A = 48 - \(\dfrac{49}{2}\) + \(\dfrac{1}{32}\)

            A  =   \(\dfrac{1536}{32}\) - \(\dfrac{784}{32}\) + \(\dfrac{1}{32}\)

             A = \(\dfrac{753}{32}\)

(1/49-1/9)(1/49-1/16).....(1/49-1/49)...(1/49-1/2401)

(1/49-1/9)(1/49-1/16).....0....(1/49-1/2401)

=0

vì khoảng cách là 7 mà 49 chia hết cho 7 nên có phân số 1/49 trong dãy

Lời giải:

$\frac{49}{5}-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{32}$

$=\frac{49}{5}-\frac{16}{32}-\frac{8}{32}-\frac{4}{32}-\frac{1}{32}$

$=\frac{49}{5}-\frac{29}{32}=\frac{1423}{160}$

a, 68 + 42 x 5 - 625 : 25

  = 68 + 210    -  25

=    278  - 25

=     253

b, 15 x { 37 + 8 + 20 } + 15 x { 13 + 22 }

  = 15 x   65               + 15 x    35

  = 15 x { 65 + 35 }

  = 15 x  100

  =  1500

c, 125 x 56 x 3

   = 7000  x 3 

   =    21000

d, 249 - { 49 - 100 }

  = 249 - 49 + 100

  = 200  + 100

  =  300

e, 1/2 + 1/4 + 1/ 8 + 1/32 + 1/64 + 1/128

   = 1/1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128

   = 1/1 - 1/128

   =   127/128

A, \(68+42\times5-625:25\)

= 68 +210 - 25

= 278 - 25

= 253

B,\(15\times\left(37+8+20\right)+15\times\left(13+22\right)\)

= \(15\times\left(37+8+20+13+22\right)\)

=\(15\times100\)

= 1500

C,\(125\times56\times3\)

= \(125\times8\times7\times3\)

=  \(1000\times21\)

= 21000

D, 249 - { 49 - 100 }

= 249 - 49 -100

= 200 - 100

= 100

E, \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)(mình sửa lại đề chút xíu nha, nó cứ bị sai sai sao ấy)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\) 

= \(1-\frac{1}{128}\)

=\(\frac{128}{128}-\frac{1}{128}\)

= \(\frac{127}{128}\)